Question

a car starts from rest and moves with constant acceleration of 4m/s² for 30 seconds. Then the brakes are applied and the car comes to rest in another 60 seconds. Find :-

i) Maximum Velocity attained by car
ii) Magnitude of retardation of car
iii) Total distance covered by the car
iv) Average speed of the car throughout the motion​

Answer 1

Answer:

1, 120 m/s

2,-2m/s^2

3,8100m

4,90m/s

Answer 2

Given:

Initial velocity, u = 0

Acceleration of the car, a = 4 m/s²

The time of acceleration, t1 = 30 s

The time of retardation, t2 = 60 s

To Find:

(i) Maximum velocity attained by car.

(ii) Magnitude of retardation of car.

(iii) Total distance covered by car.

(iv) Avg speed of car through out the motion.

Calculation:

(i) Max velocity attained by the car is given by the 1st equation of motion:

v = u + at1

⇒ v = 0 + 4 × 30

⇒ v = 120 m/s

(ii) For the retardation of car (a'):

Initial velocity will be = v = 120 m/s

Final velocity will be, v' = 0

⇒ v' = v + a' × t2

⇒ 0 = 120 + a' × 60

⇒ a' = - 2 m/s²

(iii) Total distance travelled by the car = Distance travelled during acceleration + Distance travelled during retardation

⇒ S = s1 + s2

⇒ S = (ut1 + 1/2 at1²) + (vt2 + 1/2 a't2²)

⇒ S = (0 + 1/2 × 4 × 30²) + (120 × 60 + 1/2 × (-2) × 60²)

⇒ S = 1800 + (7200 - 3600)

⇒ S = 1800 + 3600

⇒ S = 5400 m

(iv) Average speed of the car = Total distance travelled / Total time

⇒ Avg Speed = 5400/90

⇒ Avg Speed = 60 m/s

- So, Max velocity attained by car is 120 m/s; Magnitude of retardation of car is -2 m/s²; Total distance covered by car is 5400 m and Avg speed of car through out the motion is 60 m/s.