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A vehicle starts from rest and accelerates uniformly for 12 seconds to a
velocity of 10ms1. It then runs at a constant velocity and is finally came to rest
in 66m with constant . The total distance covered by the vehicle is
580m. Find the value of acceleration and the time taken.

Answer :

onyebuchinnaji

The value of the acceleration is 0.76 m/s² and the total time taken by the vehicle is 39 seconds.

Acceleration of the vehicle

The acceleration of the vehicle before coming to rest is calculated as follows;

v² = u² + 2as

where;

  • v is the final velocity
  • u is the initial velocity
  • a is the acceleration
  • s is the distance traveled before stopping

the car came to rest with constant velocity attained after 12 seconds.

the initial velocity of the car before 12 seconds is zero.

v² = 0 + 2as

a = v²/2s

a = (10²)/(2 x 66)

a = 0.76 m/s²

Time of motion of the vehicle

d = ut + ¹/₂at²

where;

  • d is the total distance traveled
  • t is the time of motion
  • a is acceleration
  • u is initial velocity of the vehicle

580 = 0 + ¹/₂(0.76)t²

580 = 0.38t²

t² = 580/0.38

t² = 1,526.3

t = √1,526.3

t = 39 seconds

Thus, the value of the acceleration is 0.76 m/s² and the total time taken by the vehicle is 39 seconds.

Learn more about time of motion here: https://brainly.com/question/2364404

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