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A car accelerates from rest at a constant rate 10 m/s2 for some time after which it decelerates at constant rate 5 m/s2 and then comes to rest. Total time elapsed is 15 s. Maximum velocity of the car is (i) while average velocity of the car during its deceleration is (ii) . Distance travelled by the car at the instant when its velocity is half of its maximum value is (iii) or (iv) .

A car accelerates from rest at a constant rate 10 m/s2 for some time after which it decelerates at constant rate 5 m/s2 and then comes to rest. Total time elapsed is 15 s. Maximum velocity of the car is (i) while average velocity of the car during its deceleration is (ii) . Distance travelled by the car at the instant when its velocity is half of its maximum value is (iii) or (iv) .

Grade:10

1 Answers

Arun
25763 Points
2 years ago
 
Let the car accelerate for t₁sec at a m/sec² and decelerate for t₂sec. at - b/ sec² 
At the end of t₁ sec. its velocity = 0 + at₁ m/sec. At the end of t₂ sec its velocity = 0 
Initial velocity at the beginning of deceleration is α t₁ 
V(t₂) = a t₁ - b t₂ = 0 Therefore t₂ = (a/ b ) t₁ ------------------------(1) 
But t = t₁ + t₂ ⇒ t = t₁ + t₂ = t₁ + (a/ b ) t₁ = t₁(1+ a/ b) 
∴ t₁ = t / (1+ a/ b) = {b / (a + b)}*t -------------------------------------- (2) 
Max. Velocity reached = a t₁ = a t / (1+ a/ b) = t / (1/ a + 1 / b) 
 = a*b*t/(a+b)
Distance covered =(a t₁² + b t₂²) /2 = ={a t₁² + b [(a/ b ) t₁)²]} /2 (substitute for t₂ from (1) 
Simplifying this = ½ * (a/ b) * (a + b)* t₁²
Now substitute for t₁ from (2) to get 
Distance covered = ½ *[a*b / (a +b)]*t²

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