Question

A car starts from rest and moves along the x-axis with constant acceleration of $5m/s^2$ 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

Initial velocity, $u=0$

Acceleration, $a=5m/s^2$

Time,

$$\begin{gathered} t_{total}=12seconds \\ {t}_1=8seconds \\ {t}_2=4seconds \end{gathered}$$

Step 2: Formula used

1. The first equation of motion, $v=u+at$.
2. The second equation of motion, $s_1=ut+\frac{1}{2}a{t}^2$.
3. $Dis\tan ce=Speed\times time$

Step 3: Determine the distance covered in the first $8seconds$

Let be the distance travelled in the first $8seconds$

Using the second equation of motion,

$$\begin{gathered} s_1=ut+\frac{1}{2}a{t}^2 \\ =u{t}_1+\frac{1}{2}a{t_1}^2 \\ =0\left(8\right)+\frac{1}{2}\left(5\right){\left(8\right)}^2 \\ =160m \end{gathered}$$$s_1$

Step 4: Determine the velocity at the end of 8 seconds when the car is accelerating

Using the first equation of motion,

$$\begin{gathered} v=u+at \\ =u+a{t}_1 \\ =0+\left(5\right)\left(8\right) \\ =40m/s \end{gathered}$$

Step 5: Determine the remaining distance

The distance traveled in first $8seconds$ has been calculated.

The total time from the moment car starts from rest is $12seconds$

Distance traveled in the last $4seconds$ will be $s_2$

$$\begin{gathered} Dis\tan ce=Speed\times time \\ {s}_2=v\times t_2 \\ =40\times 4 \\ =160m \end{gathered}$$

Step 6: Determine the total distance

$$\begin{gathered} Totaldis\tan ce=s_1+s_2 \\ =160+160 \\ =320m \end{gathered}$$

Therefore, the car covers $320m$ in total.