A car travels a distance 100 m with a constantacceleration and average

1.	A car travels a distance 100 m with a constantacceleration and average velocity of 20 m s. Thefinal velocity acquired by the car is 25 m s.Find : (i) the initial velocity and (ii) accelerationof car.Ans. (i) 15 m - (ii) 2 m s-2
Answer» Average velocity=(initial velocity + final velocity)/220 = (initial velocity + 25)/2 (1) Initial velocity = 40 - 25 = 15 m/s (2) using v = u + at Where, v = final velocity u = initial velocity t = time taken a = acceleration 25 = 15 + atat = 25 - 15 = 10.........(1) Now, using s = ut + 1/2* at^2 s = Distance travelled 100 = 15t + 1/2 10t100 = 15t + 5t20t = 100t = 100/20 = 5 s Put value of t in eq(1)at = 10a = 10/5 = 2 m/s^2 1)V (Av) =v+u/2 v+u=2V (Av)v+u=2×2025+u=40v=40-25v=15m/s The initial velocity is 15m/s 2) v^2-u^2=2as a=v^2-u^2/2s =(25^2-15^2)/(2×100) =(625-225)/200 = 400/200 =2m/s^2 The acceleration of car is 2m/s^2

A car travels a distance 100 m with a constantacceleration and average velocity of 20 m s. Thefinal velocity acquired by the car is 25 m s.Find : (i) the initial velocity and (ii) accelerationof car.Ans. (i) 15 m - (ii) 2 m s-2

Answer»

Average velocity=(initial velocity + final velocity)/220 = (initial velocity + 25)/2

(1) Initial velocity = 40 - 25 = 15 m/s

(2) using v = u + at Where, v = final velocity u = initial velocity t = time taken a = acceleration

25 = 15 + atat = 25 - 15 = 10.........(1)

Now, using s = ut + 1/2* at^2 s = Distance travelled 100 = 15t + 1/2 *10*t100 = 15t + 5t20t = 100t = 100/20 = 5 s

Put value of t in eq(1)at = 10a = 10/5 = 2 m/s^2

1)V (Av) =v+u/2 v+u=2V (Av)v+u=2×2025+u=40v=40-25v=15m/s The initial velocity is 15m/s

2) v^2-u^2=2as a=v^2-u^2/2s =(25^2-15^2)/(2×100) =(625-225)/200 = 400/200 =2m/s^2 The acceleration of car is 2m/s^2

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