a car is moving with the velocity of 339m/S the brakes applied to car produces an accleration of 6m/s2 in the opposite direction of its motion .if the car takes 2s to stop after the application of brakes ,calculate the distance travelled during this time .
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Deceleration of car is
a = (v - u) / t
= [(0 - 339) m/s] / 2 s
= -169.5 m/s²
Now,
Use equation of motion
v² - u² = 2aS
S = (v² - u²) / (2a)
= (0² - 339²) / (2 × -169.5)
= 339 m
∴ 339 m was travelled during 2 seconds
a = (v - u) / t
= [(0 - 339) m/s] / 2 s
= -169.5 m/s²
Now,
Use equation of motion
v² - u² = 2aS
S = (v² - u²) / (2a)
= (0² - 339²) / (2 × -169.5)
= 339 m
∴ 339 m was travelled during 2 seconds
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