A car starts with acceleration 2 m s-2. A scooterist starts towards the car with a velocity of 20 ms-1. If the scooterist was 96 m behind the car at the start, the time in which he overtakes the car is

Question

A car starts with acceleration 2 m s-2. A scooterist starts towards the car with a velocity of 20 ms-1. If the scooterist was 96 m behind the car at the start, the time in which he overtakes the car is (A) 8 s (B) 6 s (C) 2 s (D) cannot overtake.

Tardigrade Admin · Accepted Answer

Let t be the required time. Distance covered by car, S =(1/2) × 2 × t2=t2 Distance covered by scooter =96+ S =20 t ∴ 96+t2=20 t i.e. t2-20 t+96=0 Solving t=8 s.

Q. A car starts with acceleration $2 m s^{- 2}$ . A scooterist starts towards the car with a velocity of $20 m s^{- 1}$ . If the scooterist was $96 m$ behind the car at the start, the time in which he overtakes the car is

8 s

6 s

2 s

cannot overtake.

Let $t$ be the required time.
Distance covered by car, $S = \frac{1}{2} \times 2 \times t^{2} = t^{2}$
Distance covered by scooter $= 96 + S = 20 t$
$∴ 96 + t^{2} = 20 t$ i.e. $t^{2} - 20 t + 96 = 0$
Solving $t = 8 s$ .

Q. A car starts with acceleration 2ms−2. A scooterist starts towards the car with a velocity of 20ms−1. If the scooterist was 96m behind the car at the start, the time in which he overtakes the car is

8 s

6 s

2 s

cannot overtake.

Let t be the required time. Distance covered by car, S=21​×2×t2=t2 Distance covered by scooter =96+S=20t ∴96+t2=20t i.e. t2−20t+96=0 Solving t=8s.

Q. A car starts with acceleration $2 m s^{- 2}$ . A scooterist starts towards the car with a velocity of $20 m s^{- 1}$ . If the scooterist was $96 m$ behind the car at the start, the time in which he overtakes the car is

Let $t$ be the required time.
Distance covered by car, $S = \frac{1}{2} \times 2 \times t^{2} = t^{2}$
Distance covered by scooter $= 96 + S = 20 t$
$∴ 96 + t^{2} = 20 t$ i.e. $t^{2} - 20 t + 96 = 0$
Solving $t = 8 s$ .