Question

While coming to a stop, a car’s velocity decreases at a constant rate of -2.0 m/s2. It travels 25 m during this time. Determine the magnitude of the velocity just before it began to slow down.

Answer 1

Answer:

During the last second acceleration is −5m/s

2

and car's final velocity is zero.

Thus, from the equation,

u=v−at

u=0−(−5)×1

u=5m/s

The initial speed of the car at the start of last second is 5m/s and after one second it stops.

Therefore, the distance covered in last second will be 5m.

Answer 2

Given:-

• Final velocity ,v = 0m/s

• Acceleration ,a = -2m/s²

• Distance ,s = 25m

To Find:-

• Initial velocity ,u .

Solution:-

By using 3rd equation of motion

→ v² = u² +2as

Substitute the value we get

→ 0² = u² + 2×(-2) ×25

→ 0 = u² + (-4) × 25

→ -u² = -100

→ u² = 100

→ u = √100

→ u = 10m/s

∴ The initial velocity of the car is 10m/s.