Question

A racing car is travelling at a speed of 90 km h–1. Brakes are applied so as to produce a uniform acceleration of – 1m s-2. Find how far the train will go before it is brought to rest.​

Answer 1

$v = 0 \\ u = 90 \times \frac{5}{18} = 25ms ^{ - 1} \\ a = - 1 \\ t = 2 \\ s = ut + \frac{1}{2} ( - 1) {(2)}^{2} \\ s = 25 \times 2 - 2 \\ 50 - 2 = 48$

Explanation:

just use the 2nd derivation to find the answer

Answer 2

Answer:

The train will move 312.5 m before it is brought to rest.

Explanation:

We will use Newton's second equation of motion to solve this question. Which is given as,

$S=ut+\frac{1}{2}at^2$      (1)

Where,

S=distance traveled by the body before coming to rest

u=initial velocity of the body

t=time during which the motion occurs

a=acceleration of the body

From the question we have,

u=90km/hr=$90\times \frac{5}{18}$=25m/s

a=-1m/s²

The final velocity(v)=0    (as the train is brought to rest)

Now we need to find the time during which the motion. We will use the first equation of motion to find the time. Which is given as,

$v=u+at$        (2)

By substituting the required values in equation (2) we get;

$0=25+(-1)t$

$t=25 seconds$    (3)

By substituting all the required values in equation (1) we get;

$S=25\times 25+\frac{1}{2}\times (-1)\times (25)^2$

$S=625-312.5$

$S=312.5m$

Hence, the train will move 312.5 m before it is brought to rest.