Question

Two cars start a race with initial velocity 7ms-1 and 4ms-1 respectively.Their acceleration are 0.4ms-2 and 0.5ms-2 respectively.If the two cars finish the race at the same time,what is the distance of the track?

Answer 1

The distance of the track is 600 m

Explanation:

The two cars move by uniformly accelerated motion, so we the distance they cover after a time t is described by the following suvat equation

$d=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

For car 1, we have

$d_1 = u_1 t + \frac{1}{2}a_1 t^2$

where

$u_1 = 7 m/s$ is the initial velocity of car 1

$a_1 = 0.4 m/s^2$ is the acceleration of car 1

So the equation can be rewritten as

$d_1 = 7t + 0.2t^2$

For car 2, we have

$d_2 = u_2 t + \frac{1}{2}a_2 t^2$

where

$u_2 = 4 m/s$ is the initial velocity of car 2

$a_2 = 0.5 m/s^2$ is the acceleration of car 2

So the equation can be rewritten as

$d_2= 5t + 0.25t^2$

The two cars finish the race at the same time: this means that they cover the same distance in the same time t, so we can write

$d_1 = d_2\\7t + 0.2t^2=5t + 0.25t^2$

And solving for t, we find

$2t - 0.05t^2= 0\\t(2-0.05t)=0$

Which gives two solutions:

t = 0 (initial instant)

t = 40 s

Therefore, the distance of the track is

$d_1 = 7t +0.2t^2 = 7\cdot 40+0.2\cdot 40^2=600 m$

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