Question

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

Answer 1

$\huge\underline\mathcal\pink{Answer}$

Given:

V= 8m/s

Time= 10 s

Distance travelled= 64m

Total distance= 584m

To find:

Acceleration , Retardation and total time taken

Solution:

V=at

⇒8=a10

⇒a=0.8m/s ^2

V ^2 =U ^2 +2as

⇒U=8 ^2 +2×a64

⇒a= -64/2×64

=−0.5m/s ^2

Distance travelled= Area of graph

⇒584= 1/2×10×8+8×t ^1 + 1/2 ×16×8

⇒584=40+8t ^1+64

⇒8t^1=480

⇒t ^1 =60

Total time= 10+60+16= 86s

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Answer 2

Answer:

86s

is the answer

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