Question

A car that weighs 1.5 × 10^4 N is initially moving at a speed of 36 km/h when the brakes are applied and the car is brought to a stop in 17 m. Assuming that the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled, and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)

Answer 1

Step-by-step explanation:

Initial velocity u = 36 km/h = 10 m /s

v = 0 , accn = a , Time taken to stop = t , distance covered to stop = s

v² = u² - 2as

u² = 2as

a = u² / 2s

= 10 x 10 / 2 x 17

= 2.94 ms⁻²

Force applied = mass x acceleration

= 15000 / 9.8 x 2.94

= 4500 N

b )

v = u + at

0 = 10 - 2.94 t

t = 10 / 2.94

= 3.4 s

c )

from the relation

u² = 2as

it is clear that stopping distance is proportional to u², if acceleration a is constant .

If initial speed u is doubled , stopping distance d will become 4 times or 17 x 4 = 68 m .

d )

u = at

if a is constant time taken to stop will be proportional to initial velocity.

If initial velocity is doubled , time too will be doubled. Or time will become

3.4 x 2 = 6.8 s .