A particle moves along a straight line such t | vedclass

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

Question

(C) Average velocity $V_1$ is defined as the total displacement divided by the total time interval.Given $x(t) = \alpha t^3 + \beta t^2 + \gamma$.At $

Vedclass · Accepted Answer

$\frac{13 \alpha + 5 \beta}{27 \alpha + 6 \beta}$

Vedclass · Answer

(C) Average velocity $V_1$ is defined as the total displacement divided by the total time interval.Given $x(t) = \alpha t^3 + \beta t^2 + \gamma$.At $t = 1 \ s$,$x(1) = \alpha(1)^3 + \beta(1)^2 + \gamma = \alpha + \beta + \gamma$.At $t = 3 \ s$,$x(3) = \alpha(3)^3 + \beta(3)^2 + \gamma = 27\alpha + 9\beta + \gamma$.Displacement $\Delta x = x(3) - x(1) = (27\alpha + 9\beta + \gamma) - (\alpha + \beta + \gamma) = 26\alpha + 8\beta$.Time interval $\Delta t = 3 - 1 = 2 \ s$.$V_1 = \frac{\Delta x}{\Delta t} = \frac{26\alpha + 8\beta}{2} = 13\alpha + 4\beta$.Instantaneous velocity $V_2$ is the derivative of position with respect to time: $V(t) = \frac{dx}{dt} = 3\alpha t^2 + 2\beta t$.At $t = 3 \ s$,$V_2 = 3\alpha(3)^2 + 2\beta(3) = 27\alpha + 6\beta$.The ratio $\frac{V_1}{V_2} = \frac{13\alpha + 4\beta}{27\alpha + 6\beta}$.

Column $I$	Column $II$
$(A)$ Distance travelled in $3\,s$	$(p)$ $-20$ units
$(B)$ Displacement in $1\,s$	$(q)$ $15$ units
$(C)$ Initial acceleration	$(r)$ $25$ units
$(D)$ Velocity at $4\,s$	$(s)$ $-10$ units

Similar Questions

At time $t=0$,a car moving along a straight line has a velocity of $16 \; m/s$. It slows down with an acceleration of $a = -0.5t \; m/s^2$,where $t$ is in seconds. Mark the correct statement$(s)$.

In the $s-t$ equation $(s=10+20t-5t^2)$,match the following columns.

Column $I$ Column $II$

$(A)$ Distance travelled in $3\,s$ $(p)$ $-20$ units

$(B)$ Displacement in $1\,s$ $(q)$ $15$ units

$(C)$ Initial acceleration $(r)$ $25$ units

$(D)$ Velocity at $4\,s$ $(s)$ $-10$ units

Vedclass Test Series

Exam Paper Generator

Online Exam Module