# Vehicle on sloped road: Worked out problem

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]

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## 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 = ?

Substituting

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)

Â© BrainMass Inc. brainmass.com December 24, 2021, 4:47 pm ad1c9bdddf>https://brainmass.com/physics/work/vehicle-sloped-road-worked-problem-7554