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Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics

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”ƒ™ ƒ ‰”ƒ’Š ‘ˆ •’‡‡† ˜• •–‘’’‹‰ †‹•–ƒ ‡Ǥ Š‡ •–‘’’‹‰ †‹•–ƒ ‡ ˆ‘” –Š‡ ˆ‘—” ƒ”• ƒ– ͳͲͲ Ȁ
Š‘—”ǡ ‹• –ƒ„—Žƒ–‡† ƒ• ˆ‘ŽŽ‘™•ǣ

Car Breaking speed Stopping distance

Car 1 100 km/hr 30 meters

Car 2 100 km/hr 32 meters

Car 3 100 km/hr 34 meters

Car 4 100 km/hr 36 meters

Solution: let us now compute the stopping distance for each car. Since we seek a relationship
between velocity and distance travelled, the following equation of motion could be used.
v = U + 2as
2
2
We know that, the final velocity (v) is zero and the initial velocity is given to be 100 km/hr.

The stopping distance (s) is therefore given by:

2
0 = U + 2(−a)s we consider a to be negative as the car is decelerating.
U 2
s =
2a

Since ‘U’ is in km/hr and we have the stopping distance in meters, we multiply ‘U’ with (5/18) to
convert its units to m/s.

5U 2
18 25U 2
s = = - - - - (2.23)
2a 648 a

25U 2
Deceleration is given by: a = - - - - (2.24)
648 s

The stopping time is computed using the equation: [v = U + (-a) t]

U km S 2 5U m s 2
Since the final velocity is zero, we have: 0 = U − at = t = ∗ = t = ∗ = s
a hr m 18a s m

Page 22 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,