The x-t graph of a particle moving along a st | vedclass

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

Question

(B) The $x-t$ graph is a parabola opening downwards,which can be represented by the equation $x = at - bt^2$,where $a$ and $b$ are positive constants.

Vedclass · Accepted Answer

Vedclass · Answer

(B) The $x-t$ graph is a parabola opening downwards,which can be represented by the equation $x = at - bt^2$,where $a$ and $b$ are positive constants.The velocity $v$ is the derivative of position $x$ with respect to time $t$:$v = \frac{dx}{dt} = \frac{d}{dt}(at - bt^2) = a - 2bt$.This equation $v = a - 2bt$ represents a straight line with a negative slope $(-2b)$ and a positive intercept $(a)$.At $t = 0$,$v = a$ (positive).At $t = T$,the slope of the $x-t$ graph is zero,so $v = 0$. This gives $a - 2bT = 0$,or $a = 2bT$.At $t = 2T$,$v = a - 2b(2T) = a - 4bT = 2bT - 4bT = -2bT = -a$ (negative).Thus,the velocity decreases linearly from a positive value to zero at $t = T$ and then becomes negative,reaching $-a$ at $t = 2T$. This corresponds to the graph shown in option $B$.

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

Similar Questions

Explain instantaneous velocity.

The positions of two cars $A$ and $B$ are given by $X_A = at + bt^2$ and $X_B = ft - t^2$. At what time will both cars have the same velocity?

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 $....... \; m$.

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

The following figure shows the variation of displacement $x$ with time $t$ for a particular object. Identify the correct interpretation of the graph.

Vedclass Test Series

Exam Paper Generator

Online Exam Module