Question

A car without passenger's moving. with certain velocity on a level ground. can be stopped in a distance of 10m . if the passenger's add 25%of it's weight it's stopping distance for the same braking force and velocity is(ignore friction

Answer 1

Let the initial velocity be u.

Let the mass of the car be m.

From the equation :

v² = u² - 2as

0 = u² - 2 × 10 × a

u² = 20a

a = u²/20

Braking force = ma

= m × u²/20 = mu²/20

When the passengers are added we have the mass being :

0.25m + m = 1.25m

From the given equation :

F = ma

We can obtain the acceleration :

mu²/20 = 1.25ma

a = mu²/1.25m × 20

a = mu²/25m

a = u²/25

From the equation :

v² = u² - 2as

We do the substitution as follows :

0 = u² - 2 × u²/25 × S

U² = 2u²S/25

25U² = 2U²S

Divide through by 2u² we have :

S = 25/2

S = 12.5 m

The stopping distance = 12.5m