Question

A motorcyclist moving with constant retardation takes 10 sec and 20 to travel successive quater kilometer how much further will he travel before coming to rest

Answer 1

Let ‘u’ be the initial speed of the motorcycle when it covers the first 250 m in 10 s. Let ‘a’ be the acceleration of the motorcycle. The velocity after this 10 second is,

v = u + at

=> v = u + 10a

Now, using,

S = ut + ½ at2

=> 250 = 10u + ½ a(102)

=> 250 = 10 u + 50a

=> u + 5a = 25 …………………..(1)

For the next 20 s the, initial velocity is v (= u + 10a).

Using, S = vt/ + ½ at/2

=> 250 = (u + 10a)20 + ½ a(202)

=> 250 = 20u + 200a + 200a

=> 250 = 20u + 400a

=> 25 = 2u + 40a

=> 2u + 40a = 25 …………………(2)

(2) - 2×(1) => 2u + 40a – 2u – 10a = 25 – 50

=> 30a = -25

=> a = -25/30 = -5/6 m/s2

(1) => u + 5a = 25

=> u = 25 – 5a = 25 + 25/6 = 175/6 m/s

Let v/ be the velocity after the 20 s.

v/ = v + at/

=> v/ = (u + 10a) + a(20)

=> v/ = u + 30a

Let, S/ be the distance travelled after the 20 s before coming to rest.

0 = (v/)2 + 2aS/

=> 0 = (u + 30a)2 +2aS/

=> 0 = (175/6 – 30×5/6)2 - 2×(5/6)S/

=> 0 = (175/6 - 25)2 – (5/3)S/

=> (5/3)S/ = (25/6)2

=> (5/3)S/ = 625/36

=> S/ = 125/12 m