Name: Equilibrium of a Particle
Uploaded: 2019-07-19
Duration: 10 min 20 s
Description: How do bridges remain static even when subjected to external forces? The principle of equilibrium ensures their structural integrity. Explore the equilibrium...

Question

A car of mass

350 kg

is travelling at

30 m s^{- 1}

when it starts to slow down,

100 m

from a junction. At first, it slows just using the air resistance of

200 N

. Then, at a distance of

s m

from the junction, it slows using brakes, providing a force of

2000 N

as well as the air resistance. Find the distance from the junction at which the brakes must be applied if the car is to stop at the junction.

Accepted Answer

When the air resistance of $R_{a} = 200 N$ is acting on the car of mass $m = 350 kg$ that is travelling at $u = 30 m s^{- 1}$ it slows down.

The force acting on the car: $- R_{a} = m a \Rightarrow a = - \frac{R}{m} \Rightarrow a = - \frac{200}{350} m s^{- 2} = - \frac{4}{7} m s^{- 2}$ .

When the brakes are used it provides another resistance of $R_{b} = 2000 N$ .

Then, the force acting on the car: $- R_{b} - R_{a} = m a \Rightarrow a = \frac{m}{- R_{b} - R_{a}} = - \frac{350}{2200} m s^{- 2} = - \frac{44}{7} m s^{- 2}$ .

From the junction $s m$ the distance travelled will be $(100 - s) m$ .

At this point final velocity will be: $v^{2} = u^{2} + 2 a s \Rightarrow v = \sqrt{u^{2} + 2 a s} = \sqrt{30^{2} + 2 (- \frac{4}{7}) (100 - s)} = 900 - \frac{8}{7} (100 - s)$

Now when the brakes applied: $v^{2} = u^{2} + 2 a s \Rightarrow 0 = {(900 - \frac{8}{7} (100 - s))}^{2} + 2 (- \frac{44}{7}) s$

$\Rightarrow (- \frac{8}{7} + \frac{88}{7}) s = 900 - \frac{800}{7} \Rightarrow s = 68.75 m$ .

Hence, the distance from the junction is $68.75 m$ at which the brakes must be applied if the car is to stop at the junction.

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