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 $v$ of a particle is given by the equation $v = 6t^2 - 6t^3$,where $v$ is in $m/s$ and $t$ is time in $s$. Then:

Fill in the blanks:
$(a)$ The path length is always ............ .
$(b)$ The slope of the velocity-time graph for an object moving with acceleration is ............ .
$(c)$ If the velocity-time graph is parallel to the time-axis,the object is ............ .

$A$ particle moves along $X$-axis in such a way that its coordinate $x$ varies with time $t$ according to the equation $x = (2 - 5t + 6t^2) \ m$. The initial velocity of the particle is ....... $m/s$.

The displacement-time graph of two moving particles makes angles of $30^{\circ}$ and $45^{\circ}$ with the time axis ($X$-axis). The ratio of their velocities is

The velocity-displacement graph of a particle moving in a straight line is as shown in the figure.

Vedclass Test Series

Exam Paper Generator

Online Exam Module