Question

The velocity of a car increases from 54 km/h to 90 km/h in 5s . Find the acceleration of the car.​

Answer 1

Explanation:

u = 54km/hr

= 54×1000÷3600

= 15m/s

v = 90km/hr

= 90×1000÷3600

=25m/s

t = 5s

a= v-u÷ t

= 25-15 ÷ 5

= 10 ÷ 5

= 2m/s^2

Answer 2

Answer:

Given :-

• The velocity of a car increases from 54 km/h to 90 km/h in 5 seconds.

To Find :-

• What is the acceleration of the car.

Formula Used :-

$\bigstar$ First Equation Of Motion Formula :

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

where,

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

Solution :-

First, we have to convert initial velocity and final velocity km/hr into m/s :

✭ In case of initial velocity (u) :

$\leadsto \sf Initial\: Velocity =\: 54\: km/h$

$\leadsto \sf Initial\: Velocity =\: 54 \times \dfrac{5}{18}\: m/s\: \bigg\lgroup \sf\bold{\pink{1\: km/h =\: \dfrac{5}{18}\: m/s}}\bigg\rgroup$

$\leadsto \sf Initial\: Velocity =\: \dfrac{\cancel{270}}{\cancel{18}}\: m/s$

$\leadsto \sf\bold{\purple{Initial\: Velocity =\: 15\: m/s}}$

✭ In case of final velocity (v) :

$\leadsto \sf Final\: Velocity =\: 90\: km/h$

$\leadsto \sf Final\: Velocity =\: 90 \times \dfrac{5}{18}\: m/s\: \bigg\lgroup \sf\bold{\pink{1\: km/h =\: \dfrac{5}{18}\: m/s}}\bigg\rgroup$

$\leadsto \sf Final\: Velocity =\: \dfrac{\cancel{450}}{\cancel{18}}\: m/s$

$\leadsto \sf\bold{\purple{Final\: Velocity =\:25\: m/s}}$

Now, we have to find the acceleration :

Given :

• Initial Velocity = 15 m/s
• Final Velocity = 25 m/s
• Time Taken = 5 seconds

According to the question by using the formula we get,

$\implies \sf 25 =\: 15 + a(5)$

$\implies \sf 25 - 15 =\: 5a$

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

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

$\implies \sf 2 =\: a$

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

$\therefore$ The acceleration of the car is 2 m/s² .