Question 15
A car starts from rest and moves along the x-axis with constant acceleration $5 m s^{- 2}$ for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?

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Solution

Given, initial velocity, $u = 0$ , acceleration, $a = 5 m s^{- 2}$ and time taken, $t = 8 s$ .
Let $s_{1}$ be the distance travelled and $v$ be the final velocity.
From second equation of motion, $s = u t + \frac{1}{2} a t^{2}$
$\Rightarrow s_{1} = 0 \times 8 + \frac{1}{2} \times 5 \times 8^{2} = 160 m$
From first equation of motion, $v = u + a t \Rightarrow v = 0 + 5 \times 8 = 40 m s^{- 1}$

Now we have to find the distance covered in time $t = 12 s$
We know, $d i s t a n c e = s p e e d \times t i m e$ .
As the speed is constant, after 8 s. i.e, speed $= 40 m s^{- 1}$ , therefore in the remaining 4 s its covers a distance $s_{2} = 40 \times 4 = 160 m$
Thus total distnace covered in 12 s, $s_{1} + s_{2} = 160 + 160 = 320 m$

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Question 15 A car starts from rest and moves along the x-axis with constant acceleration 5 ms−2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?

Question 15
A car starts from rest and moves along the x-axis with constant acceleration $5 m s^{- 2}$ for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?