A race car starts from rest and accelerates uniformly to a speed of $40m{s^{ - 1}}$ in $8.0$ seconds. How far will the car travel during the $8.0$ seconds?

Hint: In order to solve this question, we will first calculate the acceleration of the car and then by using newton’s equation of motion which gives relation between distance, acceleration and initial, final velocity of an object and by using given parameters value we will solve for distance covered by the car.

Formula used:
Newton’s 3rd equation of motion is written as ${v^2} - {u^2} = 2aS$ where,
u, v is the initial, final velocity of a body.
a, S are the acceleration, distance covered by the body.
Also, acceleration ‘a’ of a body is found as $a = \dfrac{{v - u}}{t}$ where t denotes time.

Complete step by step solution:
According to the question, we have given that
$u = 0$ car starts from rest.
$v = 40m{s^{ - 1}}$ final velocity of the car after time $t = 8s$ so, by using $a = \dfrac{{v - u}}{t}$ and on putting the value of parameters, acceleration will be
$a = \dfrac{{40 - 0}}{8}$
$a = 5m{s^{ - 2}}$
Now, we have
$u = 0$ , $v = 40m{s^{ - 1}}$ and $a = 5m{s^{ - 2}}$ ,and let ‘S’ be the distance covered by the car then by using the formula ${v^2} - {u^2} = 2aS$ and on putting the values of parameter we get,
${(40)^2} - 0 = 2 \times 5 \times S$
solving for S we get,
$1600 = 10 \times S$
$ \Rightarrow S = 160m$
Hence, the distance covered by the car in $8$ seconds is $160m.$

Note:
It should be remembered that, when a body is started its motion from rest its initial velocity is always be zero and acceleration is just the rate of change of velocity of a body with respect to time and on solving newton’s equations of motion always check that all the parameters are in their respective SI units or not if not convert them in their respective SI units and then solve for asked parameter.