Question

Q] A body starts to slide over a horizontal surface with an initial velocity of 0.5 m/s. Due
to friction, its velocity decreases at the rate of 0.05 ms-2. How much distance will it t
cover in 10 seconds after start ?

Answer 1

▶ Solution :

Given :

▪ Initial velocity = 0.5m/s

▪ Retardation due to friction = 0.05m/s^2

▪ Time interval = 10s

To Find :

▪ Distance travelled by body in the given time interval.

Concept :

✒ Since, acceleration is constant throughout the motion, we can apply equation of kinematics directly.

✒ Third equation of kinematics is given by

☞ S = ut + (1/2)at^2

Calculation :

→ S = ut - (1/2)at^2

Negative sign shows retardation

→ S = (0.5×10) - (0.5×0.05×100)

→ S = 5 - 2.5

→ S = 2.5m

Answer 2

We have been given that body starts from rest at initial velocity of 0.5 m/s. Velocity decreases at the rate of 0.05 m/s²

We have to find distance covered by body after 10 seconds.

Solution

Here velocity decreases at the rate of 0.05 m/s²

So, retardation of body = 0.05 m/s²

So, acceleration of body = -0.05 m/s² [Retardation = - Acceleration ]

Using 1st equation of motion :

➼ v = u + at

➼ v = 0.5 + (-0.05 × 10)

➼ v = 0.5 - (0.5)

➼ v = 0 m/s

So, final velocity of body = 0 m/s

Again using 3rd equation of motion :

➺ v² = u² + 2as

➺ 0² = (0.5)² + 2s(-0.05)

➺ 0 = 0.25 - 0.1s

➺ 0.1s = 0.25

➺ s = 0.25/0.1

➺ s = 2.5 m

Hence, distance covered by body in 10 seconds = 2.5 m