The maximum possible acceleration of a train | vedclass

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

Question

(C) To find the minimum time,we use the velocity-time graph. The train accelerates from rest to $10\ m/s$ at $10\ m/s^2$,taking $t_1 = \frac{10}{10} =

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) To find the minimum time,we use the velocity-time graph. The train accelerates from rest to $10\ m/s$ at $10\ m/s^2$,taking $t_1 = \frac{10}{10} = 1\ s$. It then decelerates from $10\ m/s$ to rest at $5\ m/s^2$,taking $t_2 = \frac{10}{5} = 2\ s$. Let $t$ be the time for which it moves at a constant speed of $10\ m/s$. The total distance covered is the area under the $v-t$ graph: $Area = \frac{1}{2} 	imes (base_1 + base_2) 	imes height = \frac{1}{2} 	imes (t + (t + 1 + 2)) 	imes 10 = 135$. Simplifying,$5(2t + 3) = 135$,so $2t + 3 = 27$,which gives $2t = 24$,or $t = 12\ s$. The total time is $T = t_1 + t + t_2 = 1 + 12 + 2 = 15\ s$.

Similar Questions

The velocity-time graph for a particle moving along the $x$-axis is shown in the figure. The corresponding displacement-time graph is correctly shown by

The displacement-time graphs of two moving particles make angles of $30^{\circ}$ and $45^{\circ}$ with the time axis. The ratio of their velocities is

$y = (P t^2 - Q t^3) \ m$ is the vertical displacement of a ball which is moving in a vertical plane. Then the maximum height that the ball can reach is,

Two cars $A$ and $B$ start from a point at the same time in a straight line and their positions are represented by $R_{A}(t) = at + bt^2$ and $R_{B}(t) = xt - t^2$. At what time do the cars have the same velocity?

The figure shows a velocity-time graph of a particle moving along a straight line. If the particle starts from the position $x_0 = -15 \, m$,then its position at $t = 2 \, s$ will be ........ $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module