Question

a car initially at rest pick up a velocity of 72km/h in 20 seconds. if the mass of the car is 1000 kg. find 1. force develope by its engine.
2. distance covered by the car.​

Answer 1

Answer:

u=0km/h

v=72km/h=20m/s

t=20s

m=1000kg

force=ma

a=v-u/t

a=20-0/20

a=2m/s²

force=1000×2

=2000newton

distance

2as=v²-u²

2×2×s=(20)²-(0)²

4s=400

s=100m

Answer 2

Given:

• Initial velocity of Car, u = 0
• Final velocity of Car, v = 72 km/h = 72 × 5/18  m/s = 20 m/s
• Time taken for changing velocity, t = 20 sec
• Mass of Car, m = 1000 kg

To find:

• Force developed in engine, F =?
• Distance covered by Car in this time, s =?

Formula required:

• First equation of motion

       v = u + a t

• Second equation of motion

       s = u t + 1/2 a t²

• Newton's second law of motion

        F = m a

[ Where v is final velocity, u is initial velocity, t is time taken, a is acceleration, s is distance covered, F is force and m is mass ]

Solution:

Using first equation of motion calculating acceleration of Car (a =?)

→ v = u + a t

→ 20 = 0 + a ( 20 )

→ 20 = 20  a

→ a = 20 / 20

→ a = 1  m/s²

Using newton's second law of motion calculating Force developed in Engine (F =?)

→ F = m a

→ F = 1000 × 1

→ F = 1000 N

Therefore,

• Force developed in the engine would 1000 Newtons.

Now,

Using second equation of motion for calculating distance covered by Car (s =?)

→ s = u t + 1/2 a t²

→ s = ( 0 ) ( 20 ) + 1/2 ( 1 ) ( 20 )²

→ s = 1/2 × 400

→ s = 200 m

Therefore,

• Distance covered by Car would be 200 metres.