Question

an object moving with speed of 6m/s comes to rest in 10 seconds. calculate the acceleration and the distance covered by the object during this time.​

Answer 1

Answer:

Acceleration = -0.6m/s^2 , Distance Covered = 30m

Explanation:

Initial Velocity of object = u = 6m/s

Final Velocity of object = v = 0m/s        [Since, object comes to rest]

Time Taken = t = 10s

Acceleration = a = ?

Distance covered = s = ?

By First Equation Of Motion,

v = u + at

0 = 6 + a(10)

10a = -6

a = -6/10 = -0.6m/s^2

By Second  Equation Of Motion,

s = ut + 1/2at^2

s = 6(10) + 1/2(-0.6)(10)^2

s = 60 + (-30)

s = 30m

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