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Model Paper 1 — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsModel Paper 13 Marks
Question
A particle is moving along a straight line and its position is given by the relation $x=\left(t^3-6 t^2-15 t+40\right) m$ Find:
a. The time at which velocity is zero.
b. Position and displacement of the particle at that point.
c. Acceleration of the particle at that point.
✓

Answer

$
x=t^3-6 t^2-15 t+40
$
$\therefore v=\frac{d x}{d t}=\left(3 t^2-12 t-15\right) m / s$ (As velocity $=1$ st order derivative of displacement, x with respect to time, t ) and $a=\frac{d v}{d t}=(6 t-12) m / s^2$ (As acceleration = 1st order derivative of velocity, v with respect to time, t )
a. By the problem, $v =3 t ^2-12 t -15=0$
$
\begin{aligned}
& \Rightarrow 3 t^2-15 t+3 t-15=0 \\
& \Rightarrow 3 t(t-5)+3(t-5)=0 \\
& \Rightarrow(3 t+3)(t-5)=0
\end{aligned}
$
So we get, either $t=-1$ or $t=5$
But we know that time cannot be negative.
 $\therefore t =5$ seconds.
b. Now, position at $t =5 s$,
$
x_5=(5)^3-6(5)^2-15(5)+40=-60 m \text { (final position) }
$
and (ii) Now to get displacement, at $t =0 s$, position $x _0=40 m$ (initial position)
$\therefore$ Displacement from $t=0 s$ to $t =5 s$,
$
\begin{aligned}
& s=x_5-x_0 \\
& \Rightarrow s=-60-40 \\
& \Rightarrow s=-100 m
\end{aligned}
$
c. Acceleration at $t =5 s$, using the equation $a=6 t^2-12 m / s ^2$
$
\begin{aligned}
& \therefore a=6(5)-12 \\
& \Rightarrow a=(30-12) \\
& \Rightarrow a=18 m / s^2
\end{aligned}
$
This is the acceleration of the particle at that instant when velocity becomes zero.

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