Question

At an instant a traffic light turns green an automobile that has been waiting at an intersection of the road accelerates with 5m/s². At the same instant a truck travelling with a constant velocity of 20m/s overtake and passes the automobile. How far beyond the starting point will the automobile overtake the truck and how fast will it be travelling.

Answer 1

1) The car overtakes the truck at a distance of 160 m far from the intersection.

2) The velocity of the car is 40 m/s

Explanation:

1)

The car is travelling with a constant acceleration starting from rest, so its position at time t (measured taking the intersection as the origin) is given by

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

where

$a=5 m/s^2$ is the acceleration

t is the time

On the other hand, the truck is travelling at a constant velocity, therefore its position at time t is given by

$x_t(t) = vt$

where

v = 20 m/s is the velocity of the truck

t is the time

The car overtakes the truck when the two positions are the  same, so when

$x_c(t) = x_t(t)\\\frac{1}{2}at^2 = vt\\t=\frac{2v}{a}=\frac{2(20)}{5}=8 s$

So, after a time of 8 seconds. Therefore, the distance covered by the car during this time is

$x_c(8) = \frac{1}{2}(5)(8)^2=160 m$

So, the car overtakes the truck 160 m far from the intersection.

2)

The motion of the car is a uniformly accelerated motion, so the velocity of the car at time t is given by the suvat equation

$v=u+at$

where

v is the velocity at time t

u is the initial velocity

a is the acceleration

For the car in this problem, we have:

u = 0 (it starts  from rest)

$a=5 m/s^2$

And we know that the car overtakes the truck when

t = 8 s

Substituting into the equation,

$v=0+(5)(8)=40 m/s$

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