The position of a particle is given by the vector < 3.3t² , 3.2 , - 2.4t >. The magnitude of the acceleration at t=0 is 3.3 m/sec².
What is acceleration?The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration.
Given in the question The position of a particle is given by the vector
< 3.3t² , 3.2 , - 2.4t >. The magnitude of the acceleration at t=0.
Differentiating position vector for two times we get acceleration vector,
so acceleration is 3.3 m/sec²
.The position of a particle is given by the vector < 3.3t² , 3.2 , - 2.4t >. The magnitude of the acceleration at t=0 is 3.3 m/sec²
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A small box is held in place against a rough vertical wall by someone pushing on it with a force directed upward at 27 ∘ above the horizontal. The coefficients of static and kinetic friction between the box and wall are 0.40 and 0.30, respectively. The box slides down unless the applied force has magnitude 23 N .
The box slides down the wall unless an external force of magnitude 23 N is applied on it. The object is directed upward with an angle of 27° above the horizontal surface. Therefore, the mass of block is 1.90 kg
What is friction?
A friction is a kind of force which resists the sliding or rolling of objects over the surface of each other.
Applied force, F = 23 N
Coefficient of static friction, μs = 0.40
Coefficient of kinetic friction, μs = 0.30
θ = 27°
Let 'N' be the normal reaction of the wall acting on the block and 'm' be the mass of block.
Resolve the components of force 'F'
As the block is in horizontal equilibrium with the wall.
So,
F Cos27° = N
N = 23 Cos27° = 20.495 N
As the block does not slide so it means that the static friction force acting on the block balances the downwards forces (gravity) acting on the block.
The force of static friction is:
μs x N = 0.4 x 20.495 = 8.19 N .... (1)
The vertically downward force acting on the block is (mg - F Sin27°)
mg - 23 Sin 27° = mg - 10.441 ... (2)
Now by equating the forces from equation (1) and (2), we get
mg - 10.441 = 8.19
mg = 18.631
m x 9.8 = 18.631
m = 1.90 kg
Thus, the mass of block is 1.90 kg.
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Your question is incomplete, most probably the complete question is:
A small box is held in place against a rough vertical wall by someone pushing on it with a force directed upward at 27 ∘ above the horizontal. The coefficients of static and kinetic friction between the box and wall are 0.40 and 0.30, respectively. The box slides down unless the applied force has magnitude 23 N. What is the mass of the box in kilograms?
Calculate the number of moles of gas necessary to create 1.62 atm of pressure in a 4.95 L rigid container at 44.1 °C. Record your response to two decimal places.
Answer:
2.35 * 10^{-4} moles
Explanation:
T
10. An athlete swims from the north end to the south end of a 60.0m pool in
20.0s and makes the return trip to the starting position in 22.0s.
a. What is the average velocity for the first half of the swim?
b. What is the average speed for the entire trip?
c. What is the average velocity for the round trip?
a) 3 meter per second is the average velocity for the first half of the swim.
b) 2.72 meter per second is the average speed for the entire trip.
c) 2.85 meter per second is the average velocity for the round trip.
What is velocity and example?Velocity can be defined as the rate that something moves in a specific direction. as the speed of a vehicle speeding north on a highway or the pace at which a rocket lifts off. Because the velocity vector is scalar, its absolute value magnitude will always equal the motion's velocity.
Briefing:The formula to find average velocity is given by :-
Average velocity =total distance/total time taken
a) The average velocity for the first half of the swim = 60/20 =3 ms⁻¹
b) The average velocity for the entire trip of the swim=60/22=2.72 ms⁻¹
c) The average velocity for the roundtrip= 60+60/20+22 = 120/42 =2.85 ms⁻¹
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The elevator and passengers have a mass of 5500 kg. After a stop at the fifth floor, the elevator accelerates upward uniformly at 1.0 m/s2 until it is rising at 2.0 m/s. What is the magnitude of the net force acting on the elevator as it accelerates?
The net force acting on the elevator is 5940.0N
We are given that,
The acceleration of elevator upward = a =1.5m/s²
The total mass of the elevator and passenger = 5500 kg
The velocity of the elevator = v = 2.0 m/s
Therefore , the elevator accelerating upward then the net exerted force F is acting on the elevator can be calculated by the Newton's second law of motion i.e.
F = ma
Thus, the acceleration due to gravity (g) of the elevator would be taken as 9.8 m/s² then the above equation may be written as,
F = m (g + a)
If putting the values in above equation then we get the total exerted force of the elevator,
F = 5500kg ( 9.8 m/s² + 1.0m/s² )
F = 5940.0 N
Therefore, the net force acting on the elevator would be 5940.0N
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A. 0.2 m/s
B. 9.875 m/s
C. 6.2 m/s
D. 6.125 m/s
(i would like to know how to do it)
Answer:
C. 6.2 m/s
Explanation:
The magnitude of the distances are
East direction
| d1 | = 80 m
West direction
| d2 | =75m
Total magnitude of distance = 80 + 75 = 155
Total time taken = 5 + 20 = 25 s
Magnitude of speed = 155/25 = 6.2 m/s
You're pulling a suitcase through the airport on its wheels. The suitcase handle is at an angle of 62.4 to the horizontal. The pulling force you apply to the handle is 10.4 N. How much work (in Joules) are you doing on the suitcase if you pull it a distance of 25.9 m through the airport?
The work is given by:
[tex]W=Fd\cos \theta[/tex]where F is the foce acting on the object, d is the distance traveled and theta is the angle between the force and the distance. Then in this case we have:
[tex]\begin{gathered} W=(10.4)(25.9)\cos 62.4 \\ W=124.79 \end{gathered}[/tex]Therefore you are doing 124.79 J
A sled accelerates from 1.4 m/s to 7.9 m/s in 5.1 s. Determine the
acceleration of the sled.
Answer:
Explanation:
Given:
V₀ = 1.4 m/s
V = 7.9 m/s
t = 5.1 s
___________
a - ?
a = (V - V₀) / t
a = (7.9 - 1.4) / 5.1 ≈ 1.3 m/s²
Compute the wavelength in air of ultrasound with a frequency of 38 kHz if the speed of sound is 344 m/s.answer in:____ m
λ = 0.0091 m
Explanation:The frequency, f = 38 kHz
f = 38000 Hz
The speed, v = 344 m/s
The wavelength is calculated using the formula:
λ = v/f
Substitute f = 38000 Hz and v = 344 m/s into the formula and solve for λ
λ = 344/38000
λ = 0.0091 m
True or false-Normal reaction is the force
that opposes the force of gravity and acts in
the direction of the force of gravity.
O True
O False
false
normal force always acts perpendicular to the contact force
An ideal gas held at constant pressure has its temperature changed by a factor of 2.53x. By what factor does its volume increase or decrease?
In order to calculate the change in volume, we can use the relation below:
[tex]\frac{P_iV_i}{T_i}=\frac{P_fV_f}{T_f}[/tex]If the pressure remains the same and the temperature increases by 2.53x, we have:
[tex]\begin{gathered} \frac{P\cdot V_i}{T}=\frac{P\cdot V_f}{2.53T}\\ \\ V_i=\frac{V_f}{2.53}\\ \\ V_f=2.53V_i \end{gathered}[/tex]Therefore the volume also changes by a factor of 2.53.
Hello, could you help me with this question: Consider the standing wave on a guitar string and the sound wave generated by the string as a result of this vibration. What do these two waves have in common? (There may be more than one correct choice.)-They have the same amplitude.-They have the same frequency.-They have the same speed.-They have the same period.-They have the same wavelength.
Answer:
-They have the same frequency
-They have the same period.
Explanation:
The vibration of the guitar string will result in the sound wave. Then, they will have the same frequency because the first one produces the second one. If they have the same frequency, they also have the same period because frequency and period are reciprocal. Then, the answers are
-They have the same frequency
-They have the same period.
The gap between thundering and flashing of light is found to be 10 seconds.How high may the clouds be in the sky?
Answer:
Lightning and thunder are produced simultaneously, but the thunder is heard a few seconds after the lightning is seen. This is because the speed of light in air is more than the speed of sound in air.
The speed of light is given as 3×10
8
m/s. The speed of sound is 340 m/s. Thus, the entire time of 10 seconds (as mentioned in the question) between seeing the light and hearing the thunderstorm is taken by the sound to travel to the observer. Hence, the distance traveled by the thunder is given as follows.
s=vt; where,
s is the distance traveled, v is the velocity of sound and t is the time taken.
That is, 330m/s×10s=3400m=3.4km.
Thus, the distance traveled by the thunder is 3.4km.
Explanation:
Answer:
Explanation:
Given:
V = 340 m/s - Sound speed
t = 10 c
______________
S - ?
S = V·t = 340·10 = 3 400 m
Can you tell me if the answer I have gotten is correct
In order to calculate the horizontal distance the ball will travel after falling off the table, first let's calculate the time the ball will be falling, using the formula below and considering the vertical movement:
[tex]\begin{gathered} \Delta S=V_0\cdot t+a\cdot\frac{t^2}{2} \\ 1=0+9.81\cdot\frac{t^2}{2} \\ 1=4.905t^2 \\ t^2=\frac{1}{4.905} \\ t=0.45 \end{gathered}[/tex]So the ball was falling for 0.45 seconds. Let's use this time to calculate the horizontal displacement when the ball started falling:
[tex]\begin{gathered} \Delta S=V\cdot t \\ \Delta S=1.4\cdot0.45 \\ \Delta S=0.63\text{ meters} \end{gathered}[/tex]So the distance is 0.63 meters.
Poor electrical conductors have low resistance.TrueFalse
ANSWER
False
EXPLANATION
The resistance of a material is the ability of the material to oppose the flow of electrical current through it.
A conductor is a material that allows the flow of electricity through it. This implies that if a material is a poor conductor, then, it does not allow the free flow of electrical current through it. In other words, the material opposes the flow of current and can be said to have resistance.
Therefore, if a material is a poor conductor, it means that it has a high resistance.
The correct answer is false.
A football is kicked and travels 57.6 m down field. It stays in the air for a total of 2.40 seconds.
a) What is the football's initial horizontal velocity?
b) What is the football's initial vertical velocity?
c) What is the football's initial velocity magnitude?
d) At what angle was the football kicked?
e) What is the maximum height of the football?
The football's initial horizontal velocity is 24 ms⁻¹ cosθ.
Velocity is the directional speed of an object in motion as an indication of its rate of change in position as discovered from a particular frame of reference and as measured via a selected general of the time.
calculation:-
a. horizontal velocity = displacement/ time
= 57.6 m / 2.40 second
= 24 ms⁻¹ cosθ horizontal
b. Initial vertical velocity is same as horizontal velocity = 24 ms⁻¹ sinθ vertical
c. The football's initial velocity magnitude is 24 ms⁻¹
d. 24 ms⁻¹ cosθ = 57.6 m / 2.40 second
cosθ = 24/24
cosθ = 1
θ = 0° from horizontal.
e. Maximum height = U²/2g
= (24)²/2×10
= 28.8 m
Velocity is the prime indicator of the location in addition to the rapidity of the item. it may be described as the distance included through an object in unit time. speed may be defined as the displacement of the object in unit time.
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Using a lower frequency of light in a slit diffraction experiment has what effect?Select one:a.There is no difference between frequencies of light.b.fewer bright bandsc.brighter patternsd.more bright bands
Here b option is correct because the frequency affe
Quick physics problem BIG POINTS, need asap
Answer:
Explanation:
Given:
L = 68 km
V₁ = 88.2 km/h
V₂ = 97.4 km/h
_____________
t -?
Time to complete the distance for each skier:
t₁ = L / V₁ = 68 / 88.2 ≈ 0.77 h
t₂ = L / V₂ = 68 / 97.4 ≈ 0.70 h
Waiting time:
t = t₁ - t₂ = 0.77 - 0.70 = 0.07 h
or
t = 60·0.07 = 4.2 min
A student is celebrating her 17th birthday today. Mars is 1.52 times farther from the sun than Earth. How old would she be in “Martian years” if she had lived her entire life in a space colony on Mars? Round your answer to the nearest Martian year.
Answer:
9 years
Explanation:
One thing is for sure: since Mars is farther away from the sun, one martian year is greater than one Earth year. This, in turn, means that if our student lived on Mars and counted her age in terms of martian years, then she would certainly be less than 17 years old in martian years.
Now, Kepler's third law relates the orbital period ( what we call a year) to the distance from the sun.
[tex]T^2=\frac{4\pi^2}{GM}a^3[/tex]where
T = oribtal period
G = gravitational constant
M = mass of the sun
a = distance from the sun.
Now, in the case of the earth, we have
[tex]T^2_E=\frac{4\pi^2}{GM}a^3_E[/tex]where
T_E = earth's orbital period
a_E = earth's distance from the sun.
Now, in the case of mars, we know that
[tex]a_m=1.52a_E[/tex]therefore, for mars, Kepler's third law gives
[tex]T^2_m=\frac{4\pi^2}{GM}(1.52a_E)^3[/tex]which we can rewrite to get
[tex]T^2_m=\frac{4\pi^2}{GM}(1.52)^3(a_E)^3[/tex][tex]\Rightarrow T^2_m=\frac{4\pi^2}{GM}\mleft(a_E\mright)^3(1.52)^3[/tex]Now at this point remember that
[tex]T^2_E=\frac{4\pi^2}{GM}(a_E)^3[/tex]therefore, we have
[tex]T^2_m=T^2_E(1.52)^3[/tex]Taking the square root of both sides gives
[tex]T_m=\sqrt[]{T^2_E(1.52)^3}[/tex][tex]T_m=\sqrt[]{(1.52)^3}T^{}_E[/tex][tex]\boxed{T_m\approx1.874T_E\text{.}}[/tex]This tells us that one martian year is about 1.874 earth years.
Or equivalently one earth year is about 1/1.874 martian years.
[tex]T_E=\frac{1}{1.874}T_m[/tex]Therefore, if a student is 17 years old on Earth, then er equivalent age on mars would be:
[tex]undefined[/tex]
A man pushes a lawn mower on a level lawn with a force of 199 N. If 38% of the force is directed horizontally, how much work is done by the man in pushing the mower 6.1 m? answer in:____J
Given:
The applied force on the lawn mower is,
[tex]F=199\text{ N}[/tex]38% of the force is directed horizontally.
The displacement of the lawn-mower is
[tex]d=6.1\text{ m}[/tex]To find:
The work done by the man
Explanation:
The horizontal force on the lawn mower is,
[tex]\begin{gathered} F_H=F\times38\text{ \%} \\ =199\times\frac{38}{100} \\ =75.6\text{ N} \end{gathered}[/tex]The work is,
[tex]\begin{gathered} W=F_H\times d \\ =75.6\times6.1 \\ =461.2\text{ J} \end{gathered}[/tex]Hence, the work is 461.2 J.
Two
objects attract each other gravitationally
a force of 50 10 10 N when they are
apart.
If the mass of one object is
2.00 kg, what is the mass of the other
Object?
with
0.20m
M2 = 0.665 kg
The mass of other object is 0.665 kg
Solution:To solve this, we must apply Newton's law of universal gravitation, which states that the force of gravity between two massive bodies 1, 2, is proportional to their masses M1 and M2 and inversely proportional to the square of the distance d between them:
F = (GM1M2) / D²
Here G = 6.67 × 10⁻¹¹ Nm²/kg²
F = 50×10⁻¹⁰N
M1 = 2.00kg
D = .20 m
by substituting values ,
50×10⁻¹⁰N =(( 6.67 × 10⁻¹¹ Nm²/kg²)(2.00kg)M2) / (.20 m)²
50×10⁻¹⁰ = 1.334 × 10⁻¹⁰ M2 /( .20)²
(50×10⁻¹⁰) (.20)² = 1.334 × 10⁻¹⁰ M2
2 ×10⁻¹⁰ = 1.334 × 10⁻¹⁰ M2
M2 = 1.334 × 10⁻¹⁰ / 2 ×10⁻¹⁰
M2 = 0.665 kg
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A 2.4 kg block slides freely across a rough horizontal surface such that the block slows down with an acceleration of -0.8 m/s^2. What is the coefficient of kinetic friction between the block and the surface? (Use g= 10 m/s^2) *
Given:
Mass of block = 2.4 kg
Acceleration = -0.8 m/s^2
Let's ifnd the coefficient of kinetic friction,
To find the coefficient of kinteic friction, apply the formula:
[tex]\mu=\frac{F}{N}[/tex]Where:
F is frictional force
N is the normal force.
Hence, we have:
[tex]\begin{gathered} \mu=\frac{ma}{mg} \\ \\ \mu=\frac{2.4\times(0.8)}{2.4\times10} \\ \\ \mu=\frac{1.92}{24} \\ \\ \mu=0.08 \end{gathered}[/tex]Therefore, the coefficient of friction between the block and the surface is 0.08
ANSWER:
0.08
Which would require the least force to move a box along a floor with a coefficient of friction of 0.30?
Multiple Choice
pull it with a rope at an angle of 30° above the horizontal
push it at an angle of 30° below the horizontal
EITHER pull it with a rope at an angle of 30° above the horizontal OR push it at an angle of 30° below the horizontal
not enough information to solve
The least force to move a box along a floor with a coefficient of friction of 0.30 Pull it with a rope at an angle of 30° above the horizontal.
Coefficient of FrictionThe friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (μ). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force. Since both F and N are measured in units of force, the coefficient of friction is a dimensionalless quantity (such as newtons or pounds). For both static and kinetic friction, the coefficient of friction has a range of values. When an object experiences static friction, the frictional force resists any applied force, causing the object to stay at rest until the static frictional force is removed. The frictional force opposes an object's motion in kinetic friction.To learn more about Coefficient of Friction refer :
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a floor with a 0.30 coefficient of friction requires the least amount of force to move a box along it. At a 30° angle above the horizontal, pull it with a rope.
The correct answer is A.
What exactly is the friction coefficient?The resistive force of friction divided by the normal and perpendicular force (N) pushing the objects together yields the coefficient of friction (fr), which is a numerical value.
What are the two friction coefficients?The static friction coefficient and the kinetic friction coefficient are the two types of friction. Generally speaking, the coefficient of kinetic friction is lower than the coefficient of static friction.
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A metal rod 80cm long lengthens by 0.090cm when its temperature rises by 93.6 C. What is the linear expansivity of the metal?
The linear expansion of the metal is 0.000012 K⁻¹
The linear expansion of the metal rod is given by
ΔL = L₀αΔT
where,
ΔL = 0.090 cm is the linear expansion
L₀ = 80 cm is the initial length of the rod
α is the linear expansivity
ΔT = 93.6°C is the increase in temperature
By re-arranging the equation, we find the linear expansivity:
α = ΔL÷L₀ΔT = 0.090cm ÷ 80cm ×93.6K = 0.000012 K⁻¹
Temperature is the measure of hotness or coldness expressed in terms of any of several scales, together with Fahrenheit and Celsius. Temperature suggests the course in which warmness energy will spontaneously go with the flow—i.e., from a warmer body (one at a better temperature) to a less warm frame (one at a decrease temperature).
Body temperature is an early caution signal of contamination. Fever is one of the frame's first reactions to infection and is a commonplace symptom in lots of illnesses. Tracking your body temperature, can help locate disorder or contamination early.
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A beam of red light is incident at an angle of 54.3 o on an equilateral prism. If the index of refraction of red light is 1.400, at what angle does the beam emerge from the other face of the prism?
ANSWER
35.57°
EXPLANATION
Given:
• The incident angle, θ₁ = 54.3°
,• The index of refraction of red light in this prism, n₃ = 1.4
Find:
• The angle at which the beam emerges from the other face of the prism, θ₂
We have the following situation,
Using Snell's law, we can find the angle α₁,
[tex]n_1\sin\theta_1=n_3\sin\alpha_1[/tex]Solving for α₁,
[tex]\alpha_1=\sin^{-1}\left(\frac{n_1}{n_3}\sin\theta_1\right)=\sin^{-1}\left(\frac{1}{1.4}\sin54.3\degree\right)\approx35.45\degree[/tex]Now, to find the angle at which the beam emerges from the other face of the prism, we have to find angle α₂, which would be the incidence angle for the second refraction.
Let's go back to the diagram of the prism,
At the top, the beam of light forms a triangle. We know that the sum of the interior angles of any triangle is 180°. We also know that angles α₁ and α₂ are complementary to the other two interior angles of that triangle, so we have,
[tex](90-\alpha_1)+(90-\alpha_2)+60=180[/tex]Solving for α₂,
[tex]\alpha_2=90+90+60-\alpha_1-180=90+90+60-35.45-180=24.55[/tex]Now, knowing that the incidence angle at the other end of the prism is 24.55°, we can find the refraction angle using Snell's law,
[tex]n_3\sin\alpha_2=n_2\sin\theta_2[/tex]Solving for θ₂,
[tex]\theta_2=\sin^{-1}\left(\frac{n_3}{n_2}\sin\alpha_2\right)=\sin^{-1}\left(\frac{1.4}{1}\sin24.55\degree\right)\approx35.57\degree[/tex]Hence, the beam emerges from the other side of the prism at an angle of 35.57°.
A hollow sphere of mass 1.4 kg has a radius of 1.5 m and has a linear speed of 5.6 m/s at the edge. what is the angular momentum of the hollow sphere?
We will have the following:
[tex]I=(1.4kg)(1.5m)(5.6m/s)\Rightarrow I=11.76kg\cdot m^2/s[/tex]So, the moment of inertia of the hollow sphere is 11.76 kg*m^2/s.
What would be my independent,dependent variable also my control group. In the experiment the distance of the walk and run were the same 10 meters throughout
According to the data, the independent variable is distance, control group is time and dependent variables are velocity and acceleration.
23. A simple circuit has a voltage of 12V and a current of 1.8A. Find the resistance in thecircuit24. A simple circuit has a resistance of 1101 and a current of 0.9A. What is the voltagesupplied to the circuit?
23.
Given,
Voltage , V=12V
Current, I=1.8 A
To find
The resistance in the circuit
Explanation
We know
By Ohm's law,
[tex]V=RI[/tex]where R is th6e resistance
Putting the values,
[tex]\begin{gathered} 12=R\times1.8 \\ \Rightarrow R=6.67A \end{gathered}[/tex]Conclusion
The resistance is 6.67A
What is the speed of each ball when they are each 4.10 m above the ground?
ANSWER:
12.3 m/s
STEP-BY-STEP EXPLANATION:
The vertical speed can be calculated by means of the following equation:
[tex]v^2_{^{}y}=u^2_y+2\cdot a\cdot s[/tex]The speed in both cases is the same, since the angle remains the same.
We can calculate the initial speed in y since we know the angle, the value s would be the height, therefore, we substitute and calculate the final speed in y:
[tex]\begin{gathered} v^2_y=(11\cdot\sin 50)^2+2\cdot(-9.8)\cdot(-4.1) \\ v^2_y=151.36 \\ v_y=\sqrt[]{151.36} \\ v_y=12.3\text{ m/s} \end{gathered}[/tex]The speed would be 12.3 m/s in both cases
A car traveling with a speed of 64.5 mph (mi/h) What is the cars speed in meters per second ( m/s)
We will have the following:
[tex]\begin{gathered} 64.5\frac{mi}{h}\ast\frac{1609.34m}{1mi}\ast\frac{1h}{3600s}=28.83400833...m/s \\ \\ \approx28.8m/s \end{gathered}[/tex]So, the velocity given is approximately 28.8 m/s.
When Jim and Rob ride bicycles, Jim can only accelerate at three-quarters the
acceleration of Rob. Both start from rest at the bottom of a long straight road with a
constant upward slope. If Rob takes 5.0 minutes to reach the top, how much earlier
should Jim start to reach the top at the same time as Rob
The time Jim should start in order to reach the top at the same time as Rob is 1.67 minutes.
What is the acceleration of Rob?
The acceleration of Rob during the upward acceleration is calculated as follows;
v = u + at
where;
v is the constant velocity of Robu is the initial velocity = 0a is the acceleration of Robt is the time of motion of Rob = 5 mins = 300 sv = 0 + at
v = at
v = 300a ----(1)
The time of motion Jim is calculated using the following kinematic equation;
v = u + a₂t₂
where;
a₂ is acceleration of Jim = ³/₄at₂ is time of motion of Jimu is initial velocityv = 0 + (³/₄a)t₂ ----- (2)
300a = (³/₄a)t₂
300 = (³/₄)t₂
t₂ = (4 x 300) / 3
t₂ = 400 seconds = 6.67 mins
The time Jim should start in order to reach the top at the same time as Rob is calculated as follows;
Δt = t₂ - t
Δt = 6.67 mins - 5 mins
Δt = 1.67 mins
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Answer:
46s
Explanation:
displacement = (Initial Velocity)(time) + (1/2)(acceleration)(time^2)
Rob:
Initial Velocity = 0
Time = 5 mins
Acceleration = a
Displacement = (0)(5) + (1/2)(a)(5^2)
D = 12.5a
Since it’s the same road, it’s the same displacement
Jim:
Initial Velocity = 0
Time = ?
Acceleration = (3/4)a
Displacement = 12.5a
12.5a = (0)(?) + (1/2)(3/4)(a)(?^2)
12.5a = (3a/8)(?^2)
33.33 = ?^2
5.7735 = ?
Jim Time = 5.7735
Time Difference = JimTime - RobTime = 0.7735 mins
Convert to secs = 0.7735 x 60 = 46.41 secs
Answer = 46 secs
