Chapter 32 : Motion of Vehicles 665
Example 32.6. A vehicle of mass 2250 kg having its wheel base of 1·8 m is travelling
at 10 m/s on a rough plane inclined at 10° with the horizontal. The centre of gravity of the vehicle
is 1 m behind the front wheel and 90 cm above the ground.
Find the distance covered by the automobile i to stop and also time taken to do so, when the
brakes are applied on both the pairs of wheels and the vehicle is (i) going up the plane; and (ii)
coming down the plane. Take coefficient of friction offered by brakes as 0·5.
Solution. Given: Mass of vehicle (M) = 2250 kg; Wheel base = 1·8 m; Speed of the vehicle
(u) = 10 m/s; Inclination of plane (α) = 10º; Distance of the c.g. from the front wheel (x 1 ) = 1 m;
Height of the c.g. above the ground level (h) = 90 cm = 0·9 m and coefficient of friction (μ) = 0·5.
(i) Distance covered and time taken by the vehicle in coming to stop when it is going up the
plane
Fig. 32.8.
Let RF= Reaction at the front pair of wheels,
RR= Reaction at the rear pair of wheels, and
s 1 = Distance covered by the vehicle in coming to stop.
We know that kinetic energy of the vehicle before the brakes are applied=(^22) 2250 (10)
112 500 N-m
22
Mu
== ...(i)
and RF + RR=Mg cos α = 2250 × 9·8 cos 10° N
= 22 050 × 0·9848 = 21 710 N
∴ Braking resistance in both the pair of wheels,
=μ (RF + RR) = 0·5 × 21 710 = 10 855 N
and as the vehicle is going up the plane, therefore, resistance due to inclination
=Mg sin α = 2250 × 9·8 sin 10° N
= 22 050 × 0·1736 = 3828 N
∴ Total resistance = 10 855 + 3828 = 14 683 N
This total resistance stops the vehicle in a distance of s 1 metres. Therefore work done by the
force
= Force × Distance = 14 683 s 1 ...(ii)
We know that kinetic energy of 112 500 N-m [per equation (i)] of the vehicle is absorbed by
the total resistance in travelling through a distance s 1. Therefore equating eqution (i) and (ii),
112 500 = 14 683 s 1
∴ s 1 =
112 500
7.66 m
14 683
= Ans.