Question

what will be it's answer please tell​

Answer 1

Answer:

ANSWER

Given : Mass of the motorcar, M=1200kg

Initial velocity of the car, u=90 km/h=25 m/s (∵1km/h=

18

5

m/s)

Final velocity of the car, v=18 km/h=5 m/s

Time taken t=4 s

ΔP=mv−mu

ΔP=1200×5−1200×25=−24000 kg m/s

v=u+at

5=25+a×4

⟹a=−5 m/s

2

(-ve shows retardation)

∣F∣=m∣a∣=1200×5=6000 N

Hope it's helpful

Answer 2

Answer:

Mass of the motor car, m = 1200 kg

Initial velocity of the motor car, u = 90 km/h = 25 m/s

Final velocity of the motor car, v = 18 km/h = 5 m/s

Time taken, t = 4 s

According to the first equation of motion:

v = u + at

5 = 25 + a (4)

a = ˆ’ 5 m/s2

Negative sign indicates that its a retarding motion i.e. velocity is decreasing.

Change in momentum = mv ˆ’ mu = m (vˆ’u)

= 1200 (5 ˆ’ 25) = ˆ’ 24000 kg m sˆ’1

Force = Mass — Acceleration

= 1200 — ˆ’ 5 = ˆ’ 6000 N

Acceleration of the motor car = ˆ’ 5 m/s2

Change in momentum of the motor car = ˆ’ 24000 kg m sˆ’1

Hence, the force required to decrease the velocity is 6000 N.

(Negative sign indicates retardation, decrease in momentum and retarding force)