The maximum possible acceleration of a train | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To minimize the total time,the train must accelerate at its maximum rate $(a_1 = 10 \ m/s^2)$ to reach the maximum speed $(v_{max} = 10 \ m/s)$ an

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) To minimize the total time,the train must accelerate at its maximum rate $(a_1 = 10 \ m/s^2)$ to reach the maximum speed $(v_{max} = 10 \ m/s)$ and decelerate at its maximum rate $(a_2 = 5 \ m/s^2)$ to come to rest.Time taken to accelerate $(t_1)$: $v = u + a_1 t_1 \implies 10 = 0 + 10 t_1 \implies t_1 = 1 \ s$.Distance covered during acceleration $(d_1)$: $d_1 = \frac{1}{2} a_1 t_1^2 = \frac{1}{2} 	imes 10 	imes (1)^2 = 5 \ m$.Time taken to decelerate $(t_2)$: $v = u - a_2 t_2 \implies 0 = 10 - 5 t_2 \implies t_2 = 2 \ s$.Distance covered during deceleration $(d_2)$: $d_2 = v_{max} t_2 - \frac{1}{2} a_2 t_2^2 = 10 	imes 2 - \frac{1}{2} 	imes 5 	imes (2)^2 = 20 - 10 = 10 \ m$.Remaining distance to be covered at constant speed $(d_3)$: $d_3 = 135 - (d_1 + d_2) = 135 - (5 + 10) = 120 \ m$.Time taken at constant speed $(t_3)$: $t_3 = \frac{d_3}{v_{max}} = \frac{120}{10} = 12 \ s$.Total time $(T)$: $T = t_1 + t_2 + t_3 = 1 + 2 + 12 = 15 \ s$.

Similar Questions

What is stopping distance?

$A$ particle is projected with velocity $v_0$ along the $x$-axis. The deceleration of the particle is proportional to the square of the distance from the origin,i.e.,$a = -\alpha x^2$. The distance at which the particle stops is:

The velocity of a particle is $v = v_0 + gt + ft^2$. If its position is $x = 0$ at $t = 0$,then its displacement after unit time $(t = 1)$ is

The distance travelled by a particle is proportional to the square of time. Then,the particle travels with:

The relation between the position $x$ and time $t$ of a particle moving in one dimension is given by $t = \sqrt{x} + 3$,where $x$ is in meters and $t$ is in seconds. When the velocity of the particle becomes zero,the displacement of the particle is ........ $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module