. A car starts from rest and accelerates uniformly


uniformly for 10 s to a velocity of 8 ms-1 . It then runs at a constant velocity and is finally brought to rest in 64 m with constant retardation.

The total distance covered by the car is 584 m. Find the value of acceleration, retardation, and total time is taken.

 

Best Answer

Explanation 

Given: Time = 10 s

Initial velocity u = 0

Final velocity v = 8 m/s

Distance s = 64 m

Total distance covered = 584 m

To find: Value of acceleration, retardation, and total time taken.

Using first equation of motion

v = u + at

8 = a  10

Therefore, acceleration is a = 0.8 m/s2

Using third equation of motion,

v2 = u2+ 2as

u = 82+ 2  a  64

a = -64 / 2 × 6= -0.5 m/s2

Therefore retardation is 0.5 m/s2

accelerates

 

Find the total time :

Distance traveled = Area of graph

584 = 1 / 2× 10× 8+8 × t1+ 1 / 2×16 ×8

584 = 40+ 8t1+64

8t1=480

t1= 60

Total time T= 10 +60+16=86 s

Final Answer : Hence,the total time is 86 s

 

 

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