Question

A racing car has a uniform acceleration of $4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$. What distance will it cover in $10sec$ after the start?

Answer 1

Step 1. Given data:

Uniform acceleration of the racing car $\left(a\right)$= $4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$

Time $\left(t\right)=10sec$

Initial velocity $\left(u\right)=0$

Step 2. Formula used:

Motion is the phenomenon in which an object changes its position.

The first equation of motion is given as:

$v=u+at..........\left(1\right)$

Where,

$v$ = final velocity

$u$ = initial velocity

$a$ = acceleration

$t$ = time taken

The third equation of motion is given as:

$v^2-u^2=2as........\left(2\right)$

Where,

$v$ = final velocity

$u$ = initial velocity

$a$ = acceleration

$s$ = displacement

Step 3. Calculations:

Substituting all the given values in $\left(1\right)$, we get

$$\begin{gathered} v=0+4\times 10 \\ v=40\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Substitute the known parameters in the equation $\left(2\right)$ we get

$$\begin{gathered} {\left(40\right)}^2=2\times 4\times s \\ s=\frac{40\times 40}{8}=200m \end{gathered}$$

Thus, the distance will cover in $10sec$ after the start is $200m$.