Question

9 A car accelerates uniformly from 18km/h to 36km/h in 10 second. What will be the distance covered by the car in this time?.​

Answer 1

Given:-

• Initial velocity ,u = 18km/h
• Final velocity ,v = 36km/h
• Time taken ,t = 10s

To Find:-

• Distance covered by the car ,s

Solution:-

Firstly we change the unit of velocity into m/s. So we multiply the given speed by 5/18 .

Initial velocity ,u = 18×5/18 = 1×5 = 5m/s

Final velocity ,v = 36×5/18 = 2×5 = 10m/s

As we know that acceleration is defined as the rate of change in velocity.

• a = v-u/t

Substitute the value we get

$:\implies$ a = 10-5/10

$:\implies$a = 5/10

$:\implies$ a = 1/2

$:\implies$ a = 0.5m/s²

So , the acceleration of the car is 0.5m/s²

Now, we have to calculate the distance covered by the car .

Using 3rd equation of motion .

• v² = u² + 2as

where,

• v denote final velocity
• u denote initial velocity
• a denote acceleration
• s denote distance

Substitute the value we get

$:\implies$ 10² = 5² + 2×0.5 × s

$:\implies$ 100 = 25 + 1s

$:\implies$ 100 -25 = s

$:\implies$ s = 100-25

$:\implies$ s = 75m

• Hence, the distance covered by the car is 75 metres.

Answer 2

Answer:

Given :-

• A car accelerates uniformly from 18 km/h to 36 km/h in 10 seconds.

To Find :-

• What will be offline distance covered by the car in this time.

Formula Used :-

$\clubsuit$ First Equation Of Motion Formula :

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

$\clubsuit$ Third Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2as}}}$

where,

• s = Distance travelled
• u = Initial Velocity
• v = Final Velocity
• a = Acceleration
• t = Time

Solution :-

First, we have to convert km/h to m/s for both the cases (initial velocity and final velocity :

$\implies \sf Initial\: Velocity =\: 18\: km/h$

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

$\implies \sf\bold{\purple{ Initial\: Velocity =\: 5\: m/s}}$

Again,

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

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

$\implies \sf Final\: Velocity =\: 2 \times 5\: m/s$

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

Now, we have to find the acceleration of the car :

Given :

• Final Velocity = 10 m/s
• Initial Velocity = 5 m/s
• Time = 10 seconds

According to the question by using the formula we get,

$\implies \sf 10 =\: 5 + a(10)$

$\implies \sf 10 =\: 5 + 10a$

$\implies \sf 10 - 5 =\: 10a$

$\implies \sf 5 =\: 10a$

$\implies \sf \dfrac{5}{10} =\: a$

$\implies \sf 0.5 =\: a$

$\implies \sf\bold{\green{a =\: 0.5\: m/s^2}}$

Hence, the acceleration of a car is 0.5 m/s².

Again, we have to find the distance covered by the car :

Given :

• Final Velocity = 10 m/s
• Initial Velocity = 5 m/s
• Acceleration = 0.5 m/s²

According to the question by using the formula we get,

$\longrightarrow \sf (10)^2 =\: (5)^2 + 2(0.5)s$

$\longrightarrow \sf 100 =\: 25 + 2\bigg(\dfrac{5}{10}\bigg) \times s$

$\longrightarrow \sf 100 =\: 25 + \dfrac{\cancel{10}}{\cancel{10}} \times s$

$\longrightarrow \sf 100 =\: 25 + s$

$\longrightarrow \sf 100 - 25 =\: s$

$\longrightarrow \sf 75 =\: s$

$\longrightarrow \sf\bold{\red{s =\: 75\: m}}$

$\therefore$ The distance covered by the car is 75 m .