Question

A train starts from rest and acelerate at a rate of 2 m per sec square for 10 second it then runs at constant speed for 30 sec and deacelerate at
4m Per sec square at until till next station. Find the distance between them.

Answer 1

. Initial velocity u = 0 Acceleration a = 2 m/s2. Let final velocity be v (before applying breaks) t = 30 sec v = u + at ⇒ 0 + 2 × 30 = 60 m/s a) S1 = ut + ½ at2 = 900 m when breaks are applied u’ = 60 m/s v’ = 0, = 60 sec (1 min) Declaration a’ = (v-u)/t = = (0 – 60 ) / 60 = - 1 m/s2. S2 = (V^'2-u^(;2))/(2a^' ) = 1800 m Total S = S1 + S2 = 1800 + 900 = 2700 m = 2.7 km. b) The maximum speed attained by train v = 60 m/s c) Half the maximum speed = 60/2= 30 m/s Distance S = (V^2-u^2)/2a = (〖30〗^2-0^2)/(2 x 2) = 225 m from starting point When it accelerates the distance travelled is 900 m. Then again declarates and attain 30 m/s. ∴ u = 60 m/s, v = 30 m/s, a = –1 m/s2 Distance = (V^2-u^2)/2a = (〖30〗^2-〖60〗^2)/(2(-1)) = 1350 m Position is 900 + 1350 = 2250 = 2.25 km from starting point