Question

A bus starts from rest and moves with constant acceleration 8ms⁻²

At the same time, a car travelling with constant velocity 16ms⁻¹
overtakes and passes the bus. After how much time and at what distance, the bus overtakes the car?
A. t=4s,d=64m
B. t=5s,d=72m
C. t=8s,d=58m
D. None of the above

Answer 1

The bus overtakes the car after 4 seconds at a distance of 64 meters, using the equations of motion for constant acceleration and constant velocity.

Step-by-step explanation:

To determine after how much time and at what distance the bus overtakes the car, we need to use the equations of motion for constant acceleration and constant velocity.

The bus starts from rest and accelerates at 8 m/s², while the car travels at a constant velocity of 16 m/s. We set up the equations of motion for each vehicle:

• For the bus: xbus = ½ a t², where a is the acceleration of the bus.
• For the car: xcar = v t, where v is the constant velocity of the car.

To find the time t when the bus overtakes the car, we set xbus equal to xcar:

½ * 8 * t² = 16 * t

After simplifying, we get:

4 * t² = 16 * t

t² = 4 * t

t = 4 seconds (since we discard t = 0 because we are looking for the time after the bus starts moving).

Now we calculate the distance using the bus's equation of motion:

xbus = ½ * 8 * (4)² = 64 meters

Therefore, the bus overtakes the car after 4 seconds at a distance of 64 meters.