Question

8
A car moving with a speed of 15 m/s can be stopped by applying
brakes, after travelling a distance of 3m. The retardation of the car is
1) 37.5m/s² 2) 47.5m/s² 3) 57.5m/s² 4) 67.5m/s²
​

Answer 1

Answer:-

Given:-

• Initial velocity of the car (u) = 15 m s^-1
• Final velocity of the car (v) = 0 m s^-1
• Distance it covered (s) = 3 m

To Find:-

Retardation (negative acceleration) of the car.

_____________...

We know,

v² - u² = 2as

or, a = (v² - u²) / 2s

or, a = -u²/2s [Only in this case; v = 0)

where,

• v = Final velocity,
• u = Initial velocity,
• a = Acceleration and
• s = Distance.

∴ a = -(15 m/s)²/ 2(3m)

➵ a = - 225 m²/s² / 6m

➵ a = - 37.5 m s^-2 ...(Ans.)

[Note: This shows negative Acceleration. Therefore, for showing retardation, negative symbol must be erased.)

_______________...

∴ ✔ (1) 37.5 m/s²