$\text { A particle travels with velocity } v \mathrm{~ms}^{-1} \text { for } 9 \text { seconds. This is shown in the graph below. }$



1. Write down the car's velocity at t=4 .   
2 . Find the car's acceleration at t=2 .   
3. Find the total distance travelled.   

A particle moves along a straight line such that ilub g()yf 6/a hjael4i stj-+5ts velocity, v,$ \mathrm{~ms}^{-1}$ , is given by v(t)=5 t $e^{-1.2 t}$, for $ t \geq 0$ .
1. On the grid below, sketch the graph of v , for $0 \leq t \leq 3$ .

2. Find the distance travelled by the particle in the first 3 seconds.
3 . Find the maximum velocity of the particle in the first 3 seconds.

  A particle moves in a straight lij4ikpb 5z;x 2 si6rk2cne with velocity $v(t)=2 t-0.3 t^{3}+2$ , for $t \geq 0$ , where v is in $ \mathrm{ms}^{-1}$ and t in seconds.
1. Find the acceleration of the particle after 2.2 seconds.   
2. a. Find the time when the acceleration is zero.   
b. Find the velocity when the acceleration is zero.   

  A particle can move along a straight line from a point;l usy6*fb;fkpix e;q-yty-z 0uxp: 8 P . The velocity v , in 0bkp x;u8;;e yzfp6 fqty lxisu --y*:$\mathrm{m} \mathrm{s}^{-1}$ , is given by the function $v(t)=2-e^{-\sin t^{v}} $ where $t \geq 0 $ is measured in seconds.
1. Write down the first two times $ t_{1}$,$ t_{2}>0$ when the particle changes direction.      
2. a. Find the time $0\lt t \lt t_{2}$ when the particle has a maximum velocity. ≈   
b. Find the time $0\lt t \lt t_{2}$ when the particle has a minimum velocity. ≈   
3. Find the distance travelled by the particle between times $t=t_{1}$ and $t=t_{2} $.≈   

  Let $ f(t)=3 t^{2}+27$, where t>0 .
The graph of a function g is obtained when the graph of f is transformed by
a stretch by a scale of $\frac{1}{9}$ parallel to the y -axis, followed by a translation by the vector $\binom{4}{-5} $.
1. Find g(t) , giving your answer in the form $ a(t-b)^{2}+c$.  (代数式) 

A particle moves along a straight line so that its velocity in $ \mathrm{m} \mathrm{s}^{-1}$ , at time t seconds, is given by g(t) .
2. Find the distance the particle travels between t=7 and t=10 .   

  A particle moves along a straight line so that its vw1uwac0o(b+ welocity, $v \mathrm{~ms}^{-1}$ , after t seconds is given by $v(t)=1.5^{t}-4.9$ , for $ 0 \leq t \leq 6 $.
1. Find when the particle is at rest.   
2. Find the acceleration of the particle when t=3 .   
3. Find the total distance travelled by the particle.   

  The acceleration, $a \mathrm{~m} \mathrm{~s}^{-2}$ , of a particle at time t seconds is given by

$a=\frac{3}{t}+5 \cos 2 t, \text { for } t \geq 1$


The particle is at rest when t=1 .
Find the velocity of the particle when t=4 .   

A particle P starts from a poi1ofnc:z1o l+ynt O and moves along a horizontal straight line. Its vno: l+z1o1fycelocity$ v \mathrm{~ms}^{-1}$ after t seconds is given by



The following diagram shows the graph of v .


1. Find the initial velocity of particle P .
2. Find the acceleration of the particle in the first second.
3. How many times does the particle change direction in the first 8 seconds. Explain your answer.
4. Find the total distance travelled by the particle in the first 8 seconds.

  A particle moves in a straight lin3u:j cpqny 9 .*)+/bxwzv0qvj oty.x7yccnk0 e and its velocity, $v \mathrm{~ms}^{-1}$ , at time t seconds, is given by $v(t)=\left(t^{2}-2\right)^{2}$ , for $0 \leq t \leq 2$ .
1. Find the initial velocity of the particle.   
2. Find the value of t for which the particle is at rest.   
3. Find the total distance travelled by the particle in the first 2 seconds.   
4. Show that the acceleration of the particle is given by $a(t)=4 t^{3}-8 t$ .  (代数式) 
5. Find the values of t for which the velocity is positive and the acceleration is negative. a  b =   

  In this question distance is in centimetres and timeqt4ve 4so r4v7bh1l.g is in seconds.
Particle X is moving along a straight line such that its displacement from a point A , after t seconds, is given by $ s_{\mathrm{X}}=24-t-5 t^{3} e^{-0.6 t}$,$ \quad 0 \leq t \leq 30$
This is shown in the following diagram.

1. Find the value of t when particle X first changes direction.   
2. Find the total distance travelled by particle X in the first 3.5 seconds.   

Another particle, Y , moves along the same line, starting at the same time as particle X . The velocity of particle Y is given by $v_{\mathrm{Y}}=5-t, 0 \leq t \leq 30$ .
3. a. Given that particle X and Y start at the same point, find the displacement function $s_{\mathrm{Y}}$ for particle Y .  (代数式) 
b. Find the other value of t when particles X and Y meet.   

  The displacement, s , in metrv*su*qsj : d)qes, of a particle t seconds after it passes through the o)vq :jqs d*s*urigin is given by the expression $ s=\ln \left(3+t-2 e^{-t}\right), t \geq 0$ .
1. Find an expression for the velocity, v , of the particle at time t .  (代数式) 
2. Find an expression for the acceleration, a , of the particle at time t .  (代数式) 
3. Find the acceleration of the particle at time t=0 .   

  A particle P moves along a omc1aeto;-pk u2f1lf-*l sa 4straight line so that its velocity, $v \mathrm{~ms}^{-1}$ , after t seconds, is given by $v=\sin 3 t-2 \cos t-2$ , for $0 \leq t \leq 6$ . The initial displacement of P from a fixed point O is 5 metres.
1. Find the displacement of P from O after 6 seconds.   

The following sketch shows the graph of v .

2. Find when the particle is first at rest.   
3. Write down the number of times the particle changes direction.    times
4. Find the acceleration of P after 2 seconds.   
5. Find the maximum speed of P .   

  John drops a stone from the top of a cliff which is h metr3df k:.6b-mnzvcl8b p*kue n9 es above sea level. The stone strikes the water surface after 9 seconds. The velocity of the falling s lc68kz*p bnm 3:.de b-fvn9uktone, $ v \mathrm{~m} \mathrm{~s}^{-1}$, t seconds after John releases it, can be modelled by the function




1. Find the velocity of the stone when t=12 , giving your answer to the nearest $ \mathrm{m} \mathrm{s}^{-1} $.≈   
2. Calculate the value of h , giving your answer to the nearest metre.≈   

The velocity of the stone when it reaches the bottom of sea is $10 \mathrm{~m} \mathrm{~s}^{-1}$ .
3. Determine the depth of sea near the cliff, giving your answer to the nearest metre.≈   

  The displacement, s , in metres, of a particle t seconds after 9mcehh2plfy/7ia +8git passes through the h /eacl7+29g y8hpifm origin is given by the expression $s=\ln \left(1+t e^{-t}\right), t \geq 0$ .
1. Find an expression for the velocity, v , of the particle at time t .  (代数式) 
2. Find an expression for the acceleration, a , of the particle at time t .  (代数式) 
3. Find the acceleration of the particle at time t=0 .   

A particle moves along the og z*ttn14y3 wko5x:b x -axis with a velocity, $ v \mathrm{~ms}^{-1}$ , at time t seconds given by the function $v(t)=3+8 t-3 t^{2}$
For $0 \leq t \leq 4$ . The particle is initially at the origin.
1. Find the value of t when the particle reaches its maximum velocity.
2. Sketch a graph of v against t showing any points of intersection with the axes.
3 . Find the distance of the particle from the origin after 3 seconds.

  Note: In this question, distance is in metres and tim. uyxjs8:+hy ee is in seconds.
A particle P moves in a straight line for six seconds. Its acceleration during this period is given by $a(t)=-2 t^{2}+13 t-15$ , for $ 0 \leq t \leq 6$ .
1. Write down the values of t when the particle's acceleration is zero.      
2. Hence or otherwise, find all possible values of t for which the velocity of P is increasing.$a \lt t \ly b$ a =    b =   

The particle has an initial velocity of $7 \mathrm{~ms}^{-1}$ .
3. Find an expression for the velocity of P at time t .  (代数式) 
4. Find the total distance travelled by P when its velocity is decreasing.   

$\text { A particle moves in a straight line such that its velocity, } v \mathrm{~m} \mathrm{~s}^{-1} \text {, at time } t \text { seconds, is given by }$



1. Find the value of t , for t>0 , when the particle is instantaneously at rest.

The particle returns to its initial position at t=T .
2. Find the value of T . Give your answer correct to three significant figures.

A particle P moves alonkgjd9tf3u 9 p,g a straight line so that its velocity, $v, \mathrm{~ms}^{-1}$ , after t seconds, is given by $v=2 \sin t-\cos 5 t+0.1$ , for $0 \leq t \leq 4$ . The initial displacement of P from a fixed point O is 2 metres.



1. Find the displacement of P from O after 4 seconds.
2. Find the second time for t , when the particle is at rest.
3. Write down the number of times P changes direction.
4. Write down the number of times P is neither accelerating or decelerating.
5. Find the maximum distance of P from O during the time $0 \leq t \leq 4$ and justify your answer.

The velocity, $ v \mathrm{~m} \mathrm{~s}^{-1}$ , at time t seconds, of a particle moving in a straight line is given by

$\begin{array}{l}
v=\frac{\left(t^{2}-1\right) \sin t}{2} \\
\text { for } 0 \leq t \leq 2
\end{array}$

1. Determine when the particle changes direction for the first time.
2. Find the times when the acceleration of the particle is $1.4 \mathrm{~m} \mathrm{~s}^{-2}$ .
3. Find the acceleration of the particle when its speed is at its greatest.

A particle moves alongs89.mx 8ikdbgi+ cd +q the x -axis with a velocity, $ v \mathrm{~ms}^{-1}$ , at time t seconds given by the function $v(t)=2+7 t-4 t^{2} $
For $0 \leq x \leq 3$ . The particle is initially at the origin.
1. Find the value of t when the particle reaches its maximum velocity.
2. Sketch a graph of v against t showing any points of intersection with the axes.
3. Find the displacement of the particle from the origin after 2 seconds.

The displacement, in centimeters, of a particle from an or vz+lznja*38/; rhtooigin, O , at time t secondsl8/o+ j h*o; vztaz3nr, is given by $s(t)=t \sin 2 t-7 \sin t \cos t, 0 \leq t \leq 3 $.
1. Find the maximum distance of the particle from O .
2. Find the acceleration of the particle at the instant it changes direction for the the second time.

A particle moves back by* f-ndg0y v)os7 2rqand forth in a straight line. Its velocity $ v \mathrm{~m} \mathrm{~s}^{-1}$ at time t seconds is given by

$v=3 t-\frac{3}{4} t^{2}, \quad 0 \leq t \leq 7$ .

At time t=0 , the displacement s of the particle from the starting point is 1 m .
1. Find the displacement of the particle when t=5 .
2. Sketch a displacement/time graph for the particle, $0 \leq t \leq 7$ , showing clearly where the curve meets the axes and the coordinates of the points where the displacement takes the greatest and least values.

For t>7 , the displacement of the particle is given by

$s=\alpha+\beta \cos \left(\frac{2 \pi t}{7}\right)$

such that s is continuous for all $t \geq 0$ .
3. Given that s=9 when t=10.5 , find the values of $\alpha and \beta$ .
4. Find the times $t_{1}$ and $ t_{2}\left(0

  $\text { A particle moves in a straight line such that at time } t \text { seconds }(t \geq 0) \text {, its velocity is given by } v=18 t^{3} e^{-3 t^{2}} \text {. Find the exact distance travelled by the particle in the first two seconds. }$   

A particle moves such thatojkq wih 2u2r pl)-78i its velocity $v \mathrm{~m} \mathrm{~s}^{-1}$ is related to its displacement s m by the equation $v(s)=2 \arctan (\cos s), 0 \leq s \leq \pi $.
1. Find the particle's acceleration $a \mathrm{~m} \mathrm{~s}^{-2} $ in terms of s .
2. Using an appropriate graph, find the particle's displacement when its acceleration is 0.5$ \mathrm{~m} \mathrm{~s}^{-2}$ .

The acceleration, $a \mathrm{~ms}^{-2}$ of a particle moving in a vertical trajectory at time t seconds, $t \geq 0$ , is given by a(t)=-(3+v) where v is the particle's velocity in $\mathrm{ms}^{-1} $. At t=0 , the particle is at a fixed origin O and has an initial velocity of $ v_{0} \mathrm{~ms}^{-1}$ .
1. By solving an appropriate differential equation, show that the particle's velocity is given by $ v(t)=\left(v_{0}+3\right) e^{-t}-3 $.

The particle initially moves upwards until it reaches its maximum height from O , and then returns to O . Let s metres represent the particle's displacement from O , and $s_{\max } $ the maximum displacement from O .
2. 1. Show that the time T taken for the particle to reach $ s_{\max }$ satisfies the equation $ e^{-T}=\frac{3}{v_{0}+3}$ .
2. Hence, solve for T in terms of $ v_{0}$ .
3. By solving an appropriate differential equation and using the results from part (b) (i) and (ii), find an expression for $s_{\text {max }}$ in terms of $v_{0} $.

Let v(T-k) represent the particle's velocity k seconds before it reaches $ s_{\max } $, where

$v(T-k)=\left(v_{0}+3\right) e^{-(T-k)}-3$

3. By using the result from part (b) (i), show that v(T-k)=3 e^{k}-3 .

Similarly, let v(T+k) represent the particle's velocity k seconds after it reaches $ s_{\text {max }} $.
4. Deduce a similar expression for v(T+k) in terms of k .
5. Hence, show that v(T-k)+v(T+k) $\geq 0 $.