Question

A jeep is traveling at a speed of 90km/h. Brakes are applied so as to produce a uniform acceleration of -2m/s2. Find how far the jeep will go before it will come to rest. Also, find the time taken to come to rest.

Answer 1

Given :

• initial velocity of jeep, u = 90 km/h = 25 m/s
• acceleration produced due to breaks, a = -2 m/s²
• final velocity of Jeep, v = 0

To find :

• Distance covered by Jeep after applying breaks and before coming to rest, s = ?
• Time taken by Jeep to come to rest after applying breaks, t = ?

Formulae required :

• Third equation of motion

     2 a s = v² - u²

• First equation of motion

     v = u + a t

( where a is acceleration , s is distance covered , v is final velocity , u is initial velocity and t is time taken )

Solution :

Using third equation of motion

→ 2 a s = v² - u²

→ 2 (-2) s = (0)² - (25)²

→ -4 s = -625

→ 4 s = 625

→ s = 625 / 4

→ s = 156.25  m

therefore,

Distance travelled by the Jeep after applying breaks and before coming to rest is 156.25 metres.

Using first equation of motion

→ v = u + a t

→ 0 = 25 + (-2) t

→ -25 = -2 t

→ t  = 25 / 2

→ t = 12.5 s

therefore,

Time taken by Jeep to come to rest after applying breaks is 12.5 seconds.