Question

a car is travelling at a speed of 72 km/ h the breaks are applied so as to produce a uniform acceleration of -1 m/ s^2 find the distance it travels after applying brakes so that it completely stops?

Answer 1

Speed in km/hr=72

Speed in m/s=72×5/18
                     = 20m/s


v²-u²=2as
0²-(20)²=2×-1×s
-400=-2s
s=200m

∴The car will travel a distance of 200m before coming to rest.

Answer 2

Answer:

The distance car travels after applying brakes before it is brought to rest is equal to 200m.

Explanation:

Given: the initial speed of the car $u=72km/hr=20ms^{-1$

The acceleration of the car, $a=-1ms^{-2}$

Final speed of the car after comes to rest, $v=0$

From the first equation of motion:-

$v=u+at$

$(0)=(20)+(-1)t$

$-t=-20\\t=20sec$

Consider that S is the distance covered by car before it stops,

From the second equation of motion:

$S=ut+\frac{1}{2}at^{2}$                                                                 ..............(1)

Put the value of u, t and a in eq.(1);

$S=(20)(20)+\frac{1}{2}(-1)(20)^{2}$

$S=400-\frac{400}{2}$

$S=400-200$

$S=200m$

Therefore, the distance covered by the car before coming to rest is 200m.