Question

A car starts from rest and moves along the $x$-axis with constant acceleration $5\mathrm{m}{\mathrm{s}}^{-2}$ for $8seconds$. If it then continues with constant velocity, what distance will the car cover in $12seconds$ since it started from the rest?

Answer 1

Step 1:- Given data

Car Starts from rest hence

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

Acceleration $\left(a\right)=5m/s^2$

Time $\left(t\right)=8s$

Step 2: Formula used,

We know that the first equation of motion, $v=u+at$

Here $v$is the final velocity and $u$is initial velocity and $a$is acceleration and $t$is time.

Step 3: From the second equation

$$\begin{gathered} s_1=ut+\frac{1}{2}a{t}^2 \\ {s}_1=(0\times 8)+\frac{1}{2}\times 5\times {\left(8\right)}^2 \\ {s}_1=\frac{1}{2}\times 5\times {\left(8\right)}^2 \\ {s}_1=\frac{1}{2}\times 5\times 64 \\ {s}_1=5\times 32 \\ {s}_1=160m \end{gathered}$$

Step 4: Calculate the total distance,

The velocity attained in $8seconds$ is given by

$$\begin{gathered} v=u+at \\ v=0+(5\times 8) \\ v=40m/s \end{gathered}$$

The distance traveled in the next $4seconds$ is

$$\begin{gathered} s_2=40\times 4 \\ {s}_2=160m \end{gathered}$$

So, the total distance traveled by car is

$s_1+s_2=160+160=320m$

Hence, the distance will the car cover in $12seconds$ since it started from the rest is $320m.$