The velocity (v) and time (t) graph of a body | vedclass

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

Question

(A) The total distance covered is the sum of the magnitudes of the areas under the velocity-time graph.$1$. Area $A_1$ (from $t=0$ to $t=4$): This is

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) The total distance covered is the sum of the magnitudes of the areas under the velocity-time graph.$1$. Area $A_1$ (from $t=0$ to $t=4$): This is a trapezoid with parallel sides of length $4 \; s$ and $1 \; s$,and height $4 \; m/s$.$A_1 = \frac{1}{2} 	imes (4 + 1) 	imes 4 = 10 \; m$.$2$. Area $A_2$ (from $t=4$ to $t=6$): This is a triangle with base $(6 - 4) = 2 \; s$ and height $2 \; m/s$.$A_2 = \frac{1}{2} 	imes 2 	imes 2 = 2 \; m$.Total distance $= |A_1| + |A_2| = 10 + 2 = 12 \; m$.

The velocity $(v)$ and time $(t)$ graph of a body in a straight line motion is shown in the figure. The point $S$ is at $4.333 \; s$. The total distance covered by the body in $6 \; s$ is $....... \; m$.

Similar Questions

The acceleration-time graph for a particle is given. What will be the corresponding velocity-time graph?

The $x-t$ graph of a particle moving along a straight line is shown in the figure. The $a-t$ graph of the particle is correctly shown by:

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 $t=10\; s$.

Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity?

An object is moving with variable speed,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module