Question

A car starts from rest and moves with constant acceleration of 4 m/s2 for 30 seconds. Then the brakes are applied and the car comes to rest in another 60 seconds. Find [5] (i) Maximum velocity attained by car (ii) Magnitude of retardation of car (iii) Total distance covered by car (iv) Average speed of car through out the motion

Answer 1

Answer:

Explanation:I can't give answer it's so huge sorry bro

Answer 2

Given:

Initial velocity, u = 0

Acceleration of car, a = 4 m/s²

Time of acceleration, t1 = 30 s

Time of deceleration, 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, v = u + at1

⇒ v = 0 + 4 × 30

⇒ v = 120 m/s

(ii) For retardation (a'):

Initial velocity = v = 120 m/s

Final velocity, v' = 0

⇒ v' = v + a' × t2

⇒ 0 = 120 + a' × 60

⇒ a' = -2 m/s²

(iii) Total distance = 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 or 5.4 km

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

⇒ Average Speed = 5400/90

⇒ Average Speed = 60 m/s

- Hence, Maximum 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 Average speed of car through out the motion is 60 m/s.