Question

A car starts from rest and accelerates uniformly for 10 sec to a velocity of 8m/s. It then runs at a constant velocity and is finally brought to rest in 64 m with a constact retardation. The total distance covered by the car is 584 m. Findustry the value of acceleration, retardation , and total time taken

Answer 1

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Answer 2

Answer:

86 sec.

Explanation:

in case(i),

V=8m/s; U=0; t=10sec;

V=U+at = 8=0+a*10;

a=0.8m/s^2;

acceleration=0.8m/s^2.

Distance covered in this duration,

s=ut+1/2*at^2;

s=0+1/2*0.8*(10)^2;

s=40m.

in case(ii),

U=8m/s; V=0m/s; s=64m;

V^2 - U^2 = 2*a*64 = 0-64 = 128a

a = -0.5m/s^2;

retardation=0.5m/s^2.

Time taken travelling this distance,

V=U+at; 0 = 8+(-0.5)*t

t=16sec.

Total Distance covered = 40 + X + 64 = 584m

X = 584-104 = 480m.

Time taken in X(metres) = 480/8

= 60sec.

Total time taken = time taken during acceleration + in retardation + during X(metres)

= 10 + 60 + 16 = 86sec.