Question

A cyclist who starts from top of the hill accelerates uniformly with 0.5m/secsq to reach the foot with velocity of 54kmph . Find the velocity of the cyclist at the end of 5sec. Find the ratio of velocities of cyclist at the end of 21sec and 7sec

Answer 1

Given,

It is just a small step to see when the cyclist will reach to the foothills.

Initial velocity ( u ) = 0.

Final velocity ( v ) = 54 km/h = 54 m * 5 / 18 s = 15 m/s

Acceleration ( a ) = 0.5 m/s^2

Time ( t ) = ?

We have,

  v = u + at

15 m/s= 0 +  ( 0.5 m / s ^2 ) * t

15 m/s =  ( 5 m / 10 s^2 ) * t 

t = 15 m * 10 s^2 / 5 m * s

t = 30 s.

So, we concluded that the cyclist will reach the foothills in 30 s.

1. Given,

Initial velocity ( u ) = o

Time ( t ) = 5 s

Acceleration = 0.5 m / s^2

Final velocity ( v ) = ?

We have,

v = u + at

v = 0 + ( 0.5 m / s^2 ) * 5 s

v = 5 * 5 m s / 10 s^2

v = 25 m / 10 s

v = 2.5 m / s.

2. To find the ratio of velocities of cyclist at the end of 21 sec and 7 sec, we have to first find their velocities.

Given,

Initial velocity ( u ) = 0

Acceleration ( a ) = 0.5 m / s^2

Time ( t ) = 7 s.

Final velocity ( v ) = ?

v = u + at

v = 0 + ( 0.5 m / s ^2 ) * 7 s

v= 0.5 m * 7 s / s ^ 2

v = 3.5 m / s^2

Now when, Time ( t ) = 21 s.

v = u + at

v = 0 +( 0.5 m / s^2 ) * 21 s

v = 0.5 m * 21 s / s^2

v = 10.5 m / s.

Now,
         = Velocity of cyclist at the end of 21 sec : Velocity of cyclist after 7 sec

         = 10.5 m / s :  3.5 m / s

         = 3 : 1

Answer 2

Answer:

@deleted account thank you sir/mam