Question

A car accelerates uniformly from 36 km/h to 72 km/h in 5s. Calculate the acceleration and the distance covered by the car in that time. (3 marks)

Answer 1

Answer:-

Given:-

• Initial velocity (u) = 36 km/h
• Final velocity (v) = 72 km/h
• Time (t) = 5s

To Find:-

Acceleration (a) of the car.

_________...

We know,

➽ a = (v-u)/t

where,

• a = Acceleration,
• v = Final Velocity,
• u = Initial Velocity &
• t = Time taken.

Therefore,

a = [(72 - 36) km/h] / 5s

➠ a = (36 km/h)/5s

➠ a = (20 m/s)/5s

[Convert km/h to m/s by multiplying 5/18]

➠ a = 4m/s²

Therefore, acceleration was 4m/s².

#Exam_Question!

Answer 2

Initial velocity (u) = 36 km/hr
Converting it into m/Sec
So it will be 36 x 5/18 = 10 m/Sec

Final velocity (v)= 72km/hr
72km/hr = 15 m/Sec

Time = 5s

We know that
acceleration(a)= [final velocity(v)- initial velocity(u)] ÷ time
Putting the values
acceleration (a)= (15-10) ÷5
a= 5 ÷5
a= 1 m/Sec

Distance covered
Formula for distance is
s = ut + 1/2 x at^2
s -> distance
u -> initial velocity
t -> time
a-> acceleration

s = 10 x 5 + 1/2 x 1 x 5^2
s = 50 + 1/2 x 25
s = 50 + 12.5
s = 62.5 metres


Acceleration = 1m/Sec
Distance covered = 62.5 metres





(Note- Always remember to convert every value given in the same unit
Here I have converted to SI units.)