26. A vehicle of mass 120 kg is moving with auniform velocity of 108 k

1.	26. A vehicle of mass 120 kg is moving with auniform velocity of 108 km/h. The forcerequired to stop the vehicle in 10s is:(A) 120x10.8N (B) 180 N(C) 720 N(D) 360 N
Answer» M=120kgU=90km/h=10m/sV=0T=5 secA=v-u/tA=-10/5A= -2m/s2F=maF= 120×(-2)F=- 240NNegative sign shows that force must be applied in opposite direction Initial velocity ( u ) = 108 Km / h => 108 × 5 / 18 => 30 m / s ⏺️ Final velocity ( v ) = 0 ⏺️ Time ( t ) = 10 sec ✔️ We know => Acc. ( a ) = ( Final velocity - initial velocity ) / time => a = ( v - u ) / t => a = ( 0 - 30 ) / 10 => a = - 30 / 10 Here Acceleration is negative be velocity is decreasing. Now , We know force is product of mass and Acceleration , i.e => Force ( F ) = mass × acc ( a ) Mass ( m ) = 120 Kg a = - 3 m/s^2 => F = m × a => F = 120 × - 3 => Here force is negative because it is in opposite direction of motion. f fufgguhihhhhhhhzadjllr

26. A vehicle of mass 120 kg is moving with auniform velocity of 108 km/h. The forcerequired to stop the vehicle in 10s is:(A) 120x10.8N (B) 180 N(C) 720 N(D) 360 N

Answer»

M=120kgU=90km/h=10m/sV=0T=5 secA=v-u/tA=-10/5A= -2m/s2F=maF= 120×(-2)F=- 240NNegative sign shows that force must be applied in opposite direction

Initial velocity ( u ) = 108 Km / h

=> 108 × 5 / 18

=> 30 m / s

⏺️ Final velocity ( v ) = 0

⏺️ Time ( t ) = 10 sec

✔️ We know

=> Acc. ( a ) = ( Final velocity - initial velocity ) / time

=> a = ( v - u ) / t

=> a = ( 0 - 30 ) / 10

=> a = - 30 / 10

Here Acceleration is negative be velocity is decreasing.

Now , We know force is product of mass and Acceleration , i.e

=> Force ( F ) = mass × acc ( a )

Mass ( m ) = 120 Kg

a = - 3 m/s^2

=> F = m × a

=> F = 120 × - 3

Here force is negative because it is in opposite direction of motion.

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