A particle starts from rest and moves in a st | vedclass

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

Question

(A) The $v$-$t$ graph of the particle is shown in the figure.$1$. For the first part (uniform acceleration):Distance $2L = \frac{1}{2} 	imes V_m 	im

Vedclass · Accepted Answer

$\frac{6}{11}$

Vedclass · Answer

(A) The $v$-$t$ graph of the particle is shown in the figure.$1$. For the first part (uniform acceleration):Distance $2L = \frac{1}{2} 	imes V_m 	imes t_1 \implies t_1 = \frac{4L}{V_m}$$2$. For the second part (constant velocity):Distance $L = V_m 	imes t_2 \implies t_2 = \frac{L}{V_m}$$3$. For the third part (uniform retardation):Distance $3L = \frac{1}{2} 	imes V_m 	imes t_3 \implies t_3 = \frac{6L}{V_m}$$4$. Average speed $\bar{V} = \frac{	ext{Total distance}}{	ext{Total time}}$Total distance $= 2L + L + 3L = 6L$Total time $= t_1 + t_2 + t_3 = \frac{4L}{V_m} + \frac{L}{V_m} + \frac{6L}{V_m} = \frac{11L}{V_m}$$5$. Therefore,$\bar{V} = \frac{6L}{11L/V_m} = \frac{6}{11} V_m$$\frac{\bar{V}}{V_m} = \frac{6}{11}$

Similar Questions

What is the ratio of the velocity in the first $1 \ s$ to the velocity in the next $2 \ s$ based on the given displacement-time graph?

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 speed-time graph of a particle moving along a fixed direction is shown in the figure. Obtain the distance (in $m$) traversed by the particle between $t=0\; s$ and $t=10\; s$.

The variation of velocity of a particle with time moving along a straight line is illustrated in the following figure. The distance travelled by the particle in $4 \; s$ is ......... $m$.

$A$ car starts from rest and travels with uniform acceleration $\alpha$ for some time and then with uniform retardation $\beta$ and comes to rest. If the total travel time of the car is $t$,the maximum velocity attained by it is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module