Question

A car is travelling at a speed of 54 km/h. Brakes are applied so as to produce a uniform acceleration of – 0.5 m/s2. Find how far the car will go before it is brought to rest? ​

Answer 1

Answer:

given: u = 54km/hr = 54 × 5/18 = 15 m/sec

a = -0.5 m/sec², v = 0

to find distance traveled S before coming to rest

Explanation:

formula used : v² - u² = 2as

0² - 15² = 2×(-0.5)×S

=> -225 = - S

=> S = 225 m

car will travel 225 m before coming to rest

Answer 2

Answer: 225 m

Explanation:

By the third equation of motion,

$v^{2} -u^{2} =2as\\s=\frac{v^{2} -u^{2}}{2a}$

Here we have -

v = 0 (as the car is brought to rest)

u = 54 km/h = 54 *(5/18) = 15 m/s

a= -0.5 m/s

Here then,

$s=\frac{-u^{2} }{2a} =\frac{-(15^{2}) }{2*(-0.5)} \\s=\frac{-225}{-1}=225$

The car will cover 225 metres or 0.225 km before it has come to rest.