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
(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

A car starts from rest (initial velocity i.e. u is 0 m/s) and moves with a constant acceleration of 4 m/s² for 30 seconds.

{ u = 0 m/s, a = 4 m/s² and t = 30 sec }

Later, brakes are applied and the car comes to rest in another 60 seconds.

{ v = 0 m/s t' = 60 sec }

We have to find the -

(i) Maximum velocity attained by car.

(ii) The magnitude of retardation of car.

(iii) Total distance covered by a car.

(iv) The average speed of car throughout the motion.

Now,

(i) v = u + at

v = 0 + 4(30)

v = 120

Maximum velocity attained by the car is 120 m/s.

(ii) When brakes are applied and the car comes to rest in another 60 seconds.

Here we have; u' = 120 m/s, v' = 0 m/s, t' = 60 sec

{ Retardation is denoted by a' }

v' = u' + a't'

0 = 120 + a'(60)

-120 = 60a'

a' = -2

Retardation of the car is -2 m/s².

(iii) Total distance (s) covered by car = ut + 1/2 at² + u't' + 1/2 a'(t')²

= 0(30) + 1/2 (4)(30)² + 120(60) + 1/2 (-2)(60)²

= 0 + 2(900) + 7200 - 1(3600)

= 1800 + 7200 - 3600

= 5400

Total distance covered by the car is 5400 m.

(iv) Average speed = (Total distance covered by car)/(Total time taken)

Total time = t + t' = 30 + 60 = 90 sec

The average speed of the car = 5400/90 = 60 m/s.

Answer 2

♛ GIVEN :-

in case 1,

Initial velocity(u) = 0 m/s    【as the car starts from rest.】

Acceleration(a) = 4 m/s²

Time(t) = 30 s

In case 2, (when brakes were applied)

Final velocity(V) = 0 m/s      【as the car comes to rest at last.】

Initial velocity(U) = final velocity in case 1.

Time(T) = 60 s

♛ TO FIND :-

(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​

♛ SOLUTION :-

(i) We know,

v = u + at

⇒v = 0 + (4 × 30)

⇒v = 120 m/s

So, the maximum velocity attained by the car is 120 m/s.

(ii) While braking, retardation occurs.

So, U = v

⇒U = 120 m/s

V = 0 m/s

T = 60 s

Applying the formula :

V = U + AT

⇒0 = 120 + A × 60

⇒ -120 = 60A

⇒A = -2 m/s²

So, the magnitude of retardation of the car is -2 m/s².

(iii) Total distance covered =

s + S, where,

$\sf{s=ut+\frac{1}{2}at^2 }$ and $\sf{S = UT+\frac{1}{2}AT^2 }$

Putting the values :-

s + S = 0 × 30 + 1/2 × 4 × 30² + (120 × 60 + 1/2 × -2 × 60²)

⇒s + S = 2 × 30² + 7200 - (1 × 60²)

⇒s + S = 900 × 2 + 7200 - (1 × 3600)

⇒s + S = 1800 + 7200 - 3600

⇒s + S = 5400 m

So, the car covers a total distance of 5400 m or 5 km 400 m.

(iv) Average speed = (Total distance covered by car)/(Total time taken)

⇒Average speed = 5400/(30 + 60)

⇒Average speed = 5400/90

⇒Average speed = 60 m/s

So, the average speed of the car through out the motion​ is 60 m/s.