Question

A car moving with velocity of 72km/hr reduces its velocity to 54km/hr in 10s after the application of break's .Find it's acceleration.

Answer 1

Answer:

Acceleration = - 0.5 m/$s^{2}$

Explanation:

u (initial velocity) = 72 km/h

                            = 72 x 5/18 = 20 m/s

v (final velocity) = 54 km/h

                          = 54 x 5/18 =15 m/s

From first law of motion : v = u + at

a = v - u / t

  = 15 - 20 /10

  = - 0.5 m/$s^{2}$

(acceleration is negative, because the car is undergoing retardation)

Answer 2

A car moving with a velocity of 72 km/hr reduces its velocity to 54 km/hr in 10 sec.

Given that, initial velocity (u) of the car is 72 km/hr.

⇒ 72 × 5/18 = 20 m/s

Also, given that final velocity (v) of the car is 54 km/hr and time (t) is 10 sec.

⇒ 54 × 5/18 = 15 m/s

We have to find the acceleration (a) of the car.

Using First Equation of Motion,

v = u + at

→ 15 = 20 + a(10)

→ 15 - 20 = 10a

→ -5 = 10a

Divide by 10 on both sides

→ -5/10 = 10a/10

→ -1/2 = a

→ -0.5 = a

(Negative sign shows retardation or deceleration).

Therefore, acceleration of the car is -0.5 m/s².