Given acceleration function find velocity

CAIE M2 2009 June Q3

6 marks Standard +0.3

3 A particle $P$ starts from a fixed point $O$ and moves in a straight line. When the displacement of $P$ from $O$ is $x \mathrm {~m}$, its velocity is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and its acceleration is $\frac { 1 } { x + 2 } \mathrm {~m} \mathrm {~s} ^ { - 2 }$.

Given that $v = 2$ when $x = 0$, use integration to show that $v ^ { 2 } = 2 \ln \left( \frac { 1 } { 2 } x + 1 \right) + 4$.
Find the value of $v$ when the acceleration of $P$ is $\frac { 1 } { 4 } \mathrm {~m} \mathrm {~s} ^ { - 2 }$.

CAIE M2 2006 November Q7

9 marks Challenging +1.2

7 A cyclist starts from rest at a point $O$ and travels along a straight path. At time $t \mathrm {~s}$ after starting, the displacement of the cyclist from $O$ is $x \mathrm {~m}$, and the acceleration of the cyclist is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, where $a = 0.6 x ^ { 0.2 }$.

Find an expression for the velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ of the cyclist in terms of $x$.
Show that $t = 2.5 x ^ { 0.4 }$.
Find the distance travelled by the cyclist in the first 10 s of the journey.

Edexcel M3 2015 January Q1

6 marks Standard +0.8

A particle $P$ of mass 3 kg is moving along the horizontal $x$-axis. At time $t = 0 , P$ passes through the origin $O$ moving in the positive $x$ direction. At time $t$ seconds, $O P = x$ metres and the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At time $t$ seconds, the resultant force acting on $P$ is $\frac { 9 } { 2 } ( 26 - x ) \mathrm { N }$, measured in the positive $x$ direction. For $t > 0$ the maximum speed of $P$ is $32 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

Find $v ^ { 2 }$ in terms of $x$.

Edexcel M3 2019 January Q1

6 marks Standard +0.8

A particle $P$ moves on the $x$-axis. At time $t$ seconds, $t \geqslant 0$, the displacement of $P$ from the origin $O$ is $x$ metres and the acceleration of $P$ is $\left( \frac { 7 } { 2 } - 2 x \right) \mathrm { m } \mathrm { s } ^ { - 2 }$, measured in the positive $x$ direction. At time $t = 0 , P$ passes through $O$ moving with speed $3 \mathrm {~ms} ^ { - 1 }$ in the positive $x$ direction. Find the distance of $P$ from $O$ when $P$ first comes to instantaneous rest.
(6)

Edexcel M3 2022 January Q3

8 marks Challenging +1.2

A particle $P$ is moving along the $x$-axis. At time $t$ seconds, where $t \geqslant 0 , P$ is $x$ metres from the origin $O$ and is moving with speed $v \mathrm {~ms} ^ { - 1 }$

The acceleration of $P$ has magnitude $\frac { 2 } { ( 2 x + 1 ) ^ { 3 } } \mathrm {~ms} ^ { - 2 }$ and is directed towards $O$ When $t = 0 , P$ passes through $O$ in the positive $x$ direction with speed $1 \mathrm {~ms} ^ { - 1 }$

Find $v$ in terms of $x$
Show that $x = \frac { 1 } { 2 } ( \sqrt { ( 4 t + 1 ) } - 1 )$

Edexcel M3 2024 June Q3

9 marks Standard +0.8

A particle $P$ is moving along the $x$-axis.

At time $t$ seconds, where $t \geqslant 0$, the displacement of $P$ from the origin $O$ is $x$ metres and $P$ is moving with velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the positive $x$ direction. The acceleration of $P$ is $\frac { 3 \sqrt { x + 1 } } { 4 } \mathrm {~ms} ^ { - 2 }$ in the positive $x$ direction.
When $t = 0 , x = 15$ and $v = 8$

Show that $v = ( x + 1 ) ^ { \frac { 3 } { 4 } }$
Find $t$ in terms of $v$.

Edexcel M3 2007 January Q1

6 marks Moderate -0.3

A particle $P$ moves along the $x$-axis. At time $t = 0 , P$ passes through the origin $O$, moving in the positive $x$-direction. At time $t$ seconds, the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $O P = x$ metres. The acceleration of $P$ is $\frac { 1 } { 12 } ( 30 - x ) \mathrm { m } \mathrm { s } ^ { - 2 }$, measured in the positive $x$-direction.
1. Give a reason why the maximum speed of $P$ occurs when $x = 30$.
Given that the maximum speed of $P$ is $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$,
find an expression for $v ^ { 2 }$ in terms of $x$.

Edexcel M3 2013 January Q1

4 marks Moderate -0.3

A particle $P$ is moving along the positive $x$-axis. When the displacement of $P$ from the origin is $x$ metres, the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the acceleration of $P$ is $9 x \mathrm {~m} \mathrm {~s} ^ { - 2 }$.

When $x = 2 , v = 6$ Show that $v ^ { 2 } = 9 x ^ { 2 }$.
(4)

Edexcel M3 Q1

6 marks Standard +0.3

A particle $P$ moves on the positive $x$-axis. When the displacement of $P$ from $O$ is $x$ metres, its acceleration is $( 6 - 4 x ) \mathrm { m } \mathrm { s } ^ { - 2 }$, measured in the direction of $x$ increasing. Initially $P$ is at $O$ and the velocity of $P$ is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the direction $O x$.

Find the distance of $P$ from $O$ when $P$ is instantaneously at rest.

OCR M3 2013 June Q3

11 marks Standard +0.3

3 A particle $P$ of mass 0.2 kg moves on a smooth horizontal plane. Initially it is projected with velocity $0.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a fixed point $O$ towards another fixed point $A$. At time $t$ s after projection, $P$ is $x \mathrm {~m}$ from $O$ and is moving with velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, with the direction $O A$ being positive. A force of $( 1.5 t - 1 ) \mathrm { N }$ acts on $P$ in the direction parallel to $O A$.

Find an expression for $v$ in terms of $t$.
Find the time when the velocity of $P$ is next $0.8 \mathrm {~ms} ^ { - 1 }$.
Find the times when $P$ subsequently passes through $O$.
Find the distance $P$ travels in the third second of its motion.

Edexcel M3 Q3

12 marks Standard +0.8

3. A car starts from rest at the point $O$ and moves along a straight line. The car accelerates to a maximum velocity, $V \mathrm {~ms} ^ { - 1 }$, before decelerating and coming to rest again at the point $A$. The acceleration of the car during this journey, $a \mathrm {~ms} ^ { - 2 }$, is modelled by the formula $$a = \frac { 500 - k x } { 150 }$$ where $x$ is the distance in metres of the car from $O$.
Using this model and given that the car is travelling at $16 \mathrm {~ms} ^ { - 1 }$ when it is 40 m from $O$,

find $k$,
show that $V = 41$, correct to 2 significant figures,
find the distance $O A$.

OCR MEI M4 2014 June Q2

12 marks Challenging +1.8

2 On a building site, a pulley system is used for moving small amounts of material up to roof level. A light pulley, which can rotate freely, is attached with its axis horizontal to the top of some scaffolding. A light inextensible rope hangs over the pulley with a counterweight of mass $m _ { 1 } \mathrm {~kg}$ attached to one end. Attached to the other end of the rope is a bag of negligible mass into which $m _ { 2 } \mathrm {~kg}$ of roof tiles are placed, where $m _ { 2 } < m _ { 1 }$. This situation is shown in Fig. 2. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{c3ac9277-d34d-4d0e-9f9b-d0bce8c741af-2_554_711_1098_678} \captionsetup{labelformat=empty} \caption{Fig. 2}

\end{figure} Initially the system is held at rest with the rope taut, the counterweight at the top of the scaffolding and the bag of tiles on the ground. When the counterweight is released, the bag ascends towards the top of the scaffolding. At time $t \mathrm {~s}$ the velocity of the counterweight is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ downwards. The counterweight is made from a bag of negligible mass filled with sand. At the moment the counterweight is released, this bag is accidentally ripped and after this time the sand drops out at a constant rate of $\lambda \mathrm { kg } \mathrm { s } ^ { - 1 }$.

Find the equation of motion for the counterweight while it still contains sand, and hence show that $$v = g t + \frac { 2 g m _ { 2 } } { \lambda } \ln \left( 1 - \frac { \lambda t } { m _ { 1 } + m _ { 2 } } \right) .$$
Given that the sand would run out after 10 seconds and that $m _ { 2 } = \frac { 4 } { 5 } m _ { 1 }$, find the maximum velocity attained by the counterweight towards the ground. You may assume that the scaffolding is sufficiently high that the counterweight does not hit the ground before this velocity is reached.

CAIE M2 2016 June Q3

6 marks Standard +0.8

A particle $P$ of mass $0.4$ kg is released from rest at a point $O$ on a smooth plane inclined at $30°$ to the horizontal. When the displacement of $P$ from $O$ is $x$ m down the plane, the velocity of $P$ is $v \text{ ms}^{-1}$. A force of magnitude $0.8e^{-5x}$ N acts on $P$ up the plane along the line of greatest slope through $O$.

Show that $v \frac{dv}{dx} = 5 - 2e^{-x}$. [2]
Find $v$ when $x = 0.6$. [4]

CAIE Further Paper 3 2022 June Q3

5 marks Challenging +1.2

A particle $P$ is moving in a horizontal straight line. Initially $P$ is at the point $O$ on the line and is moving with velocity $25 \text{ m s}^{-1}$. At time $t$ s after passing through $O$, the acceleration of $P$ is $-\frac{4000}{(5t + 4)^3} \text{ m s}^{-2}$ in the direction $PO$. The displacement of $P$ from $O$ at time $t$ is $x$ m. Find an expression for $x$ in terms of $t$. [5]

Edexcel M3 2005 January Q5

12 marks Standard +0.8

At time $t = 0$, a particle $P$ is at the origin $O$, moving with speed 18 m s$^{-1}$ along the $x$-axis, in the positive $x$-direction. At time $t$ seconds ($t > 0$) the acceleration of $P$ has magnitude $\frac{3}{\sqrt{(t + 4)}}$ m s$^{-2}$ and is directed towards $O$.

Show that, at time $t$ seconds, the velocity of $P$ is $[30 - 6\sqrt{(t + 4)}]$ m s$^{-1}$. [5]
Find the distance of $P$ from $O$ when $P$ comes to instantaneous rest. [7]

Edexcel M3 2011 January Q1

6 marks Standard +0.3

A particle $P$ moves on the positive $x$-axis. When the distance of $P$ from the origin $O$ is $x$ metres, the acceleration of $P$ is $(7 - 2x)$ m s$^{-2}$, measured in the positive $x$-direction. When $t = 0$, $P$ is at $O$ and is moving in the positive $x$-direction with speed 6 m s$^{-1}$. Find the distance of $P$ from $O$ when $P$ first comes to instantaneous rest. [6]