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Question

A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t then, the distance travelled by the car is

A
12(αβα+β)t2
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B
12(α+βαβ)t2
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C
12(α2+β2αβ)t2
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D
12(α2β2αβ)t2
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Solution

The correct option is A 12(αβα+β)t2
Lets consider, v= maximum velocity,
When A car accelerates from rest at a constant rate α
From 1st equation of motion,
v=u+at
v=0+αt1
v=αt1
t1=vα
After car decelerate at the rate β,
From 1st equation of motion,
v=u+at
0=vβt2
v=βt2
t2=vβ
Total time elapsed,
t=t1+t2
t=vα+vβ
t=v(α+βαβ)
v=(αβα+β)t
From the above figure,
The total distance covered is equal to the area under the velocity-time graph
S=12×v×t
s=12(αβα+β)t2
The correct option is A.

1539181_1192056_ans_2f7baca0db8f412d8eda411bb974ba58.jpeg

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