Physic Tutorials

# How to solve questions on Kinematics and deformation of solids

How to solve questions on Kinematics and deformation of solids

Question 1

A train with an initial velocity of 20ms-1 is subjected to a uniform deceleration of 2ms-2. The time required to bring the train to a complete halt is
A. 40s         B. 5s            C. 10s          D. 20s

Solution

initial velocity = 20 ms-1

final velocity = 0 ms-1 ( since the train completely come to a halt)

deceleration = 2ms-1

V = u + at

0 = 20 + (-2) * t

-20 = -2t

t = 20/2  = 10 s

C is the correct option

Question 2

Calculate the work done, when a force of 20N stretches a spring by 50mm.
A. 2.5 J      B. 0.5 J         C. 1.5 J        D. 2.0 J

Solution

workdone = 1/2 f e

f = 20 N

e = 50 mm = 0.05 m

workdone = 1/2 * 20 * 0.05   =  0.5 J

B is the correct option

Question 3 and 4 are from UTME 2012 Physics Questions – Type Yellow

Question 3

A car starts from rest and moves with a uniform acceleration of 30ms-2 for 20s. Calculate the distance covered at
the end of the motion.
A. 6km        B. 12km         C. 18km            D. 24km.

Solution

initial velocity = 0 ms-1 ( since it is starting from rest)

acceleration = 30 ms-1

time taken = 20 s

using

s = ut + 1/2at^2

s =0 + 1/2 * 30 * 20^2

s = 400*15

s = 6000 m  = 6 km

A is the correct option

Question 4

If a load of 1kg stretches a cord by 1.2cm, what is the force constant of
the cord?
A. 866 Nm-1          B. 833 Nm-1           C. 769 Nm-1             D. 667 Nm-1

Solution

Mass = 1kg

weight = mg = 1 * 10 = 10 N

extension = 1.2 cm  = 0.012 m

force constant = f / e

k = 10 / 0.012

k = 833 Nm-1

B is the correct option

Question 5

This question is from advance physics Nelkon and parker

A car moving with a velocity of 36 kmh-1 accelerates uniformly at 1 ms-2 until it reaches a velocity of 54 kmh-1. Calculate

i the time taken

ii the distance travelled during the acceleration

Solution

x kmh-1 to ms-1 = (x *1000) / 3600

36 kmh-1 = (36 * 1000) / 3600 = 10 ms-1

54 kmh-1 = (54 * 1000) / 3600 = 15 ms-1

initial velocity = 10 ms-1

final velocity = 15 ms-1

v = u + at

15 = 10 + 1t

t = 15 – 10

t = 5 s

ii

v^2 = u^2 + 2as

15^2 = 10^2 + 2*1*s

225 = 100 + 2s

2s = 225 – 100

2s = 125

s = 62.5 m

Recommended: Short note on Kinematics

### Bolarinwa Olajire

A lecturer, Educationist, PhD student at FUNAAB, and a Blogger.