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

The position $x$ of a particle at any instant is related to its velocity $v$ by the equation $v = \sqrt{2x + 9}$. The particle starts from the origin. What are the initial acceleration and initial velocity of the particle?

Starting from rest,the acceleration of a particle is given by $a = 2(t - 1)$. The velocity of the particle at $t = 5 \, s$ is ......... $m/s$.

If the velocity of a particle is $v = At + Bt^2$,where $A$ and $B$ are constants,then the distance travelled by it between $1 \ s$ and $2 \ s$ is

$A$ car is moving with velocity $V$. It stops after applying brakes at a distance of $20 \ m$. If the velocity of the car is doubled,then what distance will it cover after applying the brakes (in $m$)?

$A$ car moving with a velocity $6.25 \,ms^{-1}$ is decelerated with $2.5 \sqrt{v} \,ms^{-2}$ (where $v$ is the instantaneous velocity). The time taken by the car to come to rest is: (in $\,s$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module