Question

A Scooter acquires a velocity 36km/h
In 10 Seconds Just after the start. it
takes 20 Seconds to stop. Calculate the
Acceleration in the two cases.​

Answer 1

Answer:

Given :-

• A scooter acquires a velocity 36 km/h in 10 seconds just after the start and it takes 20 seconds to stop.

To Find :-

• What is the acceleration in both cases.

Formula Used :-

$\clubsuit$ First Equation Of Motion Formula :

$\longmapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time

Solution :-

${\normalsize{\bold{\purple{\underline{\bigstar\: In\: first\: case\: :-}}}}}$

First, we have to convert final velocity km/h to m/s :

$\implies \sf Final\: Velocity =\: 36\: km/h$

$\implies \sf Final\: Velocity =\: 36 \times \dfrac{5}{18}\: m/s$

$\implies \sf Final\: Velocity =\: \dfrac{180}{18}\: m/s$

$\implies \sf\bold{\green{Final\: Velocity =\: 10\: m/s}}$

Given :

• Final Velocity = 10 m/s
• Initial Velocity = 0 m/s
• Time = 10 seconds

According to the question by using the formula we get,

$\longrightarrow \sf 10 =\: 0 + a(10)$

$\longrightarrow \sf 10 =\: 0 + 10a$

$\longrightarrow \sf 10 - 0 =\: 10a$

$\longrightarrow \sf 10 =\: 10a$

$\longrightarrow \sf \dfrac{\cancel{10}}{\cancel{10}} =\: a$

$\longrightarrow \sf 1 =\: a$

$\longrightarrow \sf\bold{\red{a =\: 1\: m/s^2}}$

$\therefore$ The acceleration in the first case is 1 m/s².

$\rule{150}{2}$

${\normalsize{\bold{\purple{\underline{\bigstar\: In\: second\: case\: :-}}}}}$

First, we have to convert initial velocity km/h to m/s :

$\implies \sf Initial\: Velocity =\: 36\: km/h$

$\implies \sf Initial\: Velocity =\: 36 \times \dfrac{5}{18}$

$\implies \sf Initial\: Velocity =\: \dfrac{180}{18}$

$\implies \sf\bold{\green{Initial\: Velocity =\: 10\: m/s}}$

Given :

• Initial Velocity = 10 m/s
• Final Velocity = 0 m/s
• Time = 20 seconds

According to the question by using the formula we get,

$\longrightarrow \sf 0 =\: 10 + a(20)$

$\longrightarrow \sf 0 =\: 10 + 20a$

$\longrightarrow \sf 0 - 10 =\: 20a$

$\longrightarrow \sf - 10 =\: 20a$

$\longrightarrow \sf \dfrac{- 1\cancel{0}}{2\cancel{0}} =\: a$

$\longrightarrow \sf \dfrac{- 1}{2} =\: a$

$\longrightarrow \sf - 0.5 =\: a$

$\longrightarrow \sf\bold{\red{a =\: -\: 0.5\: m/s^2}}$

As we see that the acceleration is - 0.5 m/s².

So, it is a retardation.

$\therefore$ The acceleration in second case is 0.5 m/s².

[Note : Retardation is a negetive acceleration.

Retardation is a acceleration with a negetive sign. ]