A lorry travelling on the level road at 40 km/h can be stopped by its brakes in a distance 16 m. Find the speed from which it can be brought to rest in the same distance when descending a hill whose angle of slope is Sin^-1 [1/15]© BrainMass Inc. brainmass.com December 24, 2021, 4:47 pm ad1c9bdddf
SOLUTION This solution is FREE courtesy of BrainMass!
The speed of the lorry = 40 km/h
= 40 x 5/18 m/s = 100/9 m/s
Distance in which it is stopped = 16 m
Using, V^2 = U^2 + 2aS to find the retardation 'a'
V = 0, U=(100/9) m/s, S = 16m, and a = ?
0 = (100/9)^2 + 2 a x 16
from this, a = -3.86m/s^2
This is the acceleration on the wheels when break is applied
(note that it is negative)
Therefore force of brakes F = M a = 3.86xM Newton
where M is the mass of the lorry
When the lorry is descending the slope, the gravitational force has a component mgSin(A)
The resultant force on the lorry = 3.86M - (9.8/15)M
= 3.21 M Newton
hence retardation = 3.21 m/s^2
Let U1 be the velocity required
we can write,
0 = U1^2 - 2 x 3.21 x 16
U1 = [2 x 3.21 x 16]^(1/2)
= 10.14 m/s
You can convert it to Km/h by multiplying with (18/5)
To convert km/h to m/s, multiply by (5/18)© BrainMass Inc. brainmass.com December 24, 2021, 4:47 pm ad1c9bdddf>