Question

5 m/s2'retederation is given for a body moving with a velocity 40m/s.
a) find the time taken by the body to come to stop?
b) what is the displacement in this time?

Answer 1

Answer:

(i):      8 s

(ii):     160 m

Explanation:

Acceleration(a) = - retardation

                         = -5 m/s²

Initial velocity(u) = 40 m/s

     As the car finally stops(due to this retardation),  final velocity(v) = 0

   Using the equations of motion:

(i):      v = u + at

        0 = 40 + (-5)(t)

       8 = t

Time taken to stop is 8 s.

(ii):     v² = u² + 2aS

        0² = 40² + 2(-5)S

        10S = 1600

        S = 160 m

magnitude of displacement is 160m

Answer 2

Given

• Final velocity (v) = 0m/s

• Initial velocity (u) = 40m/s

• Acceleration (a) = 5m/s²

Negative sign means retardation

To find

• Time taken to stop

• Displacement (s)

According to the question it by using Newton equation of motion where

• v = Final velocity

• u = Initial velocity

• t = Time

• a = Acceleration

Substituting the values

0 = 40 + (-5) × t

0 = -40 = -5 t

- 40 = - 5

40 ÷ 5 = t

8 = t

So, by the body to store of his 8 s.

According to the question in by using

Newton's 2nd equation of motion where

S = it+½ at²

Where

S = Distance

U = Initial velocity

A = Acceleration

T = Time

Substituting the value

S = 40 × 8 + ½ × ( -5 ) × 8 × 8

S = 320 + ( -5 ) × 4 × 8

S = 320 - 160

S = 160

So, the distance is 160 minutes.