Question

A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a constant acceleration of 2.10m/s2, and the car has an acceleration of 3.40m/s2. The car overtakes the truck after the truck has moved 60.0 m.
(A) How much time does it take the car to overtake the truck?
(B) How far was the car behind the truck initially?
(C) What is the speed of each when they are abreast?
(D) On a single graph, sketch the position of each vehicle as a function of time. Take x = 0 at the initial location of the truck.

Answer 1

A) The car overtakes the truck after 7.56 s

B) Initial distance between car and truck: 37.1 m

C) Speed of the truck: 15.9 m/s, speed of the car: 25.7 m/s

D) See graph in attachment

Explanation:

A)

The truck starts from rest and has a constant acceleration, so its position at time t can be written as

$x_t(t)=d+\frac{1}{2}a_tt^2$

where

d is the initial distance between the truck and the car (the truck starts some distance ahead of the car)

$a_t=2.10 m/s^2$ is the acceleration of the truck

The car position instead it is given by the equation

$x_c(t)=\frac{1}{2}a_ct^2$

where

$a_c=3.40 m/s^2$ is the acceleration of the car

The car overtakes the truck when the truck has moved 60.0 m, so when

$x_t(t') = d + 60$

Therefore, solving the equation, we find the time t when  this occurs:

$d+\frac{1}{2}a_t t'^2 = d+60\\\frac{1}{2}a_tt'^2=60\\t'=\sqrt{\frac{2\cdot 60}{a_t}}=\sqrt{\frac{120}{2.1}}=7.56 s$

B)

In order to find the initial distance between the car and the truck (d), we have to calculate first the distance covered by the car during these 7.56 s. It is given by:

$x_c(t')=\frac{1}{2}a_c t'^2=\frac{1}{2}(3.40)(7.56)^2=97.2 m$

This means that after 7.56 s, when the car reaches the truck, the car has covered 97.2 m while the truck has covered 60 m. However, their positions are now equal, so we can write:

$x_c(t')=x_t(t')$

And by solving the equation, we find the value of d, the initial distance between car and truck:

$\frac{1}{2}a_c t'^2 = d + \frac{1}{2}a_t t'^2\\d=\frac{1}{2}(a_c-a_t)t'^2 = \frac{1}{2}(3.40-2.10)(7.56)^2=37.1 m$

C)

In order to find the speed of each vehicle, we use the following suvat equation:

$v=u+at$

where

u is the initial velocity

a is the acceleration

t is the time

For the truck, we have:

u = 0

$a_t = 2.10 m/s^2$

So its speed after t = 7.56 s is

$v_t = 0+(2.10)(7.56)=15.9 m/s$

For the car, we have

u = 0

$a_c=3.40 m/s^2$

So its speed after t = 7.56 s is

$v_c=0+(3.40)(7.56)=25.7 m/s$

D)

Find the graph required in attachment.

On the x-axis, it is represented the time in seconds. On the y-axis, it is represented the position in meters.

Both curves are in the shape of a parabola since the motion of both vehicles is an accelerated motion.

The curve that starts at -37.1 m is the curve representing the car: in fact, the car starts behind the truck by 37.1 m. The curve that starts from x = 0, t= 0 is that of the truck.

The two curves meets when t = 7.56 s: at that time, the two vehicles have reached the same position, and we see that occurs when x = 60 m, which means that this happens when the truck has covered 60 meters.

Learn more about accelerated motion:

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