$A$ particle moves along a straight line such that its displacement $x$ varies with time $t$ as $x = \alpha t^3 + \beta t^2 + \gamma$, where $\alpha, \beta, \gamma$ are constants. $V_1$ is the average velocity of the particle during its journey between $t = 1 \ s$ and $t = 3 \ s$. $V_2$ is the instantaneous velocity of the particle at $t = 3 \ s$. The ratio $\frac{V_1}{V_2}$ is
- A$\frac{27 \alpha + 9 \beta}{26 \alpha + 6 \beta}$
- B$\frac{9 \alpha + 3 \beta}{18 \alpha + 4 \beta}$
- C$\frac{13 \alpha + 5 \beta}{27 \alpha + 6 \beta}$
- D$\frac{26 \alpha + 8 \beta}{9 \alpha + 3 \beta}$
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Similar Questions
For any arbitrary motion in space,which of the following relations are true?
$(a)$ $v_{\text{average}} = (1/2) (v(t_1) + v(t_2))$
$(b)$ $v_{\text{average}} = [r(t_2) - r(t_1)] / (t_2 - t_1)$
$(c)$ $v(t) = v(0) + at$
$(d)$ $r(t) = r(0) + v(0)t + (1/2)at^2$
$(e)$ $a_{\text{average}} = [v(t_2) - v(t_1)] / (t_2 - t_1)$
(The 'average' stands for the average of the quantity over the time interval $t_1$ to $t_2$.)
$(a)$ $v_{\text{average}} = (1/2) (v(t_1) + v(t_2))$
$(b)$ $v_{\text{average}} = [r(t_2) - r(t_1)] / (t_2 - t_1)$
$(c)$ $v(t) = v(0) + at$
$(d)$ $r(t) = r(0) + v(0)t + (1/2)at^2$
$(e)$ $a_{\text{average}} = [v(t_2) - v(t_1)] / (t_2 - t_1)$
(The 'average' stands for the average of the quantity over the time interval $t_1$ to $t_2$.)
Medium
View Solution$A$ body is at rest at $x=0$. At $t=0$,it starts moving in the positive $x-$direction with a constant acceleration. At the same instant,another body passes through $x=0$ moving in the positive $x$ direction with a constant speed. The position of the first body is given by $x_{1}(t)$ after time $t$ and that of the second body by $x_{2}(t)$ after the same time interval. Which of the following graphs correctly describes $(x_{1}-x_{2})$ as a function of time $t$?
MediumAIEEE 2008
View SolutionIn 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
| 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 |
Medium
View SolutionWhen the velocity of a body is variable,then:
Easy
View SolutionThe relation between time $t$ and distance $x$ is $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. The relation between acceleration $a$ and velocity $v$ is:
DifficultJEE MAIN 2024
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