Question

A bike starts from rest and accelerates with 3 m/s2 for 10 seconds, then come to rest in next 5 seconds, by uniform retardation.Then find the total distance travelled before coming to rest

Answer 1

Answer:

The bike had covered the distance of150m

Answer 2

Given: The initial velocity will be = 0 m/s

           Acceleration = 3 ms⁻²

           Time taken = 10 seconds

           After this time, the body comes to rest in the next 5 seconds by uniform retardation.

To find: The total distance traveled before coming to rest

Solution:

Distance covered in the first 10 seconds will be calculated by the formula,

S = ut + 1/2at² ( Where S is the distance, u is the initial velocity, a is the acceleration, t is the time taken )

Putting the value in equation,

S  = 0×10 + 1/2×3×10×10

   = 0 + 150 = 150 m

The final velocity will be calculated by the formula, v = u + at

v = 0 + 3×10 = 10 m/s

This final velocity will become the initial velocity for the second phase.

Retardation produced by the body = v' - u'/t'

a' = 0 - 10/5 = - 2ms⁻²

Now, the body will undergo retardation and will come to rest. So, the distance traveled by the body will be calculated by S' = u't' + 1/2a't'²

S' = 10×5 + 1/2 (-2)(5)²

   = 150 - 25

   = 125 m

The total distance traveled b the body before coming to rest will be equal to S' + S = 150 + 125 = 275 m

Therefore, the distance traveled = 275 m.