Question

Upon signal of starting the race an athlete starts from rest with a constant acceleration of 8m/s2. At the same time a cyclist travelling with a constant speed of 72km/h passes the athlete. How far beyond the starting point will the athlete overtake the cyclist and what will be the speed of the athlete at the time he overtakes the cyclist?

Answer 1

here u go with the answer

Answer 2

Answer:

Option A) 100m and 40 m/s.

Explanation:

The acceleration of the athlete is given in the question as 8 m/s^2. So, we know that to find the distance we have to use the formulae of S=ut +1/2gt^2 since, initial velocity u is 0 m/s. So, S= 1/2*8*t^2 or S=4t^2.

The cyclist is traveling with 72km/h so distance S of cyclist will be 72*5/18*t or 20t. No, at time of overtake the distance will be same so 4t^2=20t, on solving we will get the value of the t = 5 seconds.

So, the distance for the athlete to overtake the cyclist will be 4t^2 or 4*5^2 which will be 100m.

So, v = u + at or now the final velocity will be v = 8*5 = 40 m/s.