Question

A car running with a velocity 72 kmph on a level road, is stopped after travelling a distance of 30 m after disengaging it's engine. If the car has a stopping distance of 80 m on cement road then k is (g=10 m/s2):

Answer 1

Intial velocity of car, u = 54 km/h = 54 × 5/18 m/s
u = 15 m/s

Car stops on applying brakes 20m
Displacement is covered by car ,S = 20m
finally car stops ∴ final velocity of car , v = 0

Now, use formula
V² = u² + 2aS , here a is Acceleration
⇒ 0 = (15)² + 2 × a × 20
⇒ -225 = 40a
⇒ a = -5.625 m/s² [ here negative sign shows that acceleration is opposite direction of motion of car ]

Hence, magnitude of acceleration = 5.625 m/s²

hope it helps you!!

Answer 2

Intial velocity of car, u = 54 km/h = 54 × 5/18 m/s

u = 15 m/s

Car stops on applying brakes 20m

Displacement is covered by car ,S = 20m

finally car stops ∴ final velocity of car , v = 0

Now, use formula

V² = u² + 2aS , here a is Acceleration

⇒ 0 = (15)² + 2 × a × 20

⇒ -225 = 40a

⇒ a = -5.625 m/s² [ here negative sign shows that acceleration is opposite direction of motion of car ]

Hence, magnitude of acceleration = 5.625 m/s²

Read more on Brainly.in - https://brainly.in/question/7838283#readmore