Kinematics

Kinematics

Definitions

Displacement is the position of a point relative to a fixed origin O. It is a vector. The SI Unit is the metre (m). Other metric units are centimeter (cm), kilometer (km). It can be positive , negative or zero.

Velocity is the rate of change of displacement with respect to time. It is a vector. The SI Unit is metre per second (ms^-1). Other metric units include kmh^-1 .

Uniform velocity is the constant speed in a fixed direction.

Speed is the magnitude of the velocity and is a scalar quantity. It is the rate of change of its distance with respect to time.

Acceleration is the rate of change of velocity with respect to time. It is a vector. The SI Unit is metre per second square (ms^-2 ). Other metric units include kmh^-2 .

Negative acceleration is also referred to as retardation. Uniform acceleration is the constant acceleration in a fixed direction.

A particle is instantaneously at rest when its velocity is zero.

A particle reaches maximum velocity when its acceleration is zero.

Formulas

 $Average\ velocity=\frac{change\ in\ displacement}{time\ taken}$  
 $Average\ speed=\ \frac{total\ dis\tan ce\ travelled}{time\ taken}$  
 $Acceleration=\frac{change\ in\ velocity}{time\ taken}$  

Graphs in Kinematics 

Displacement - time graphs

A displacement-time graph for a body moving in a straight line shows its displacement, x, from a fixed point on the line plotted against time, t.
The velocity, v, of the body at time, t, is given by the gradient of the graph since  $\frac{dx}{dt}=v$  .

The displacement-time graph for a body moving with constant velocity is a straight line. The velocity, v, of the body is given by the gradient of the line.

The displacement-time graph for a body moving with variable velocity is a curve.

The velocity at any time, t, may be estimated from the gradient of the tangent to the curve at that time. The average velocity between two times may be estimated from the gradient of the chord joining them. 

Velocity - time graphs

Velocity-time graph for a body moving in a straight line shows its velocity, v, plotted against time, t.

The acceleration, a, of a body at time, t, is given by the gradient of the graph at t, since  $a=\frac{dv}{dt}$  .

The displacement in a time interval is given by the area under the velocity-time graph for that time interval.

The velocity-time graph for a body moving with uniform acceleration is a straight line.

The acceleration of the body is given by the gradient of the line. The velocity-time graph for a body moving with variable acceleration is a curve. 

Variable motion of a Particle

Application of Differentiation
Consider a particle moving in a straight line and its displacements is a function of t, that is s = f(t).
since s = f(t) then its derivative  $\frac{ds}{dt}=f'\left(t\right)\ which\ is\ the\ rate$  of change of s with respect to t. 

So the velocity v at time t is given by :  $v=\frac{ds}{dt}=f'\left(t\right)$  .

 $a=\frac{dv}{dt}=\frac{d^2s}{dt^2}=f''\left(t\right)$  
At a maximum, velocity = 0.
 At rest ds/dt = 0.

2018 P2 8

(b) A bus starts from rest at Station A and travels a distance of 80km in 60mins to station B. Since the bus arrived at Station B early, it remained there for 20mins then started the journey to station C. Thee time taken to travel from Station B to Station C was 90 minutes at an average speed of 80kmh^-1.

(i) Draw a distance-time graph to illustrate the motion of the bus.

(ii) Determine the distance from station B to Station C .

(iii) Determine the average speed from Station A to station B, in kmh^-1.

(ii) Average speed = 30km per HOUR
time taken = 90 mins = 1.5 hrs
Average speed =  $\frac{dis\tan ce\ travelled}{time\ taken}$  

 $80=\frac{x}{1.5}$  
 $x=80\times1.5\ =120km$  

(ii) 60 mins = 1 hr 
Average speed =  $\frac{80}{1}=80kmh^{-1}$  

2018 P2 8

(a) A particle moves in a straight line so that its distance, s metres , after t seconds, measured from a fixed point, O, is given by the function  $s=t^3-2t^2+t-1$  . 

Determine 
(i) its velocity when t = 2
(ii) the values of t when the particle is at rest
(iii) the distance between the rest points 
(iv) the time at which the maximum velocity occurs. 

(i)  $\frac{ds}{dt}=v$                                      ( iii) $when\ t=1$                                    (iv)  $at\ the\ \max\ ,\ v=0$  

 $v=3t^2-4t+1$                            $s=\left(1\right)^3-2\left(1\right)^2+1-1$               $3t^2-4t+1=0$  
 $when\ t=2$                                       $=1-2+1-1$                                $\left(t-1\right)\left(3t-1\right)=0$  
 $v=3\left(2\right)^2-4\left(2\right)+1$                    $=-1\ =1$                                                  $t=1\ or\ t=\frac{1}{3}$                                                                                     
 $=5ms^{-1}$                                                                                             $\max\ velocity\ occurs\ at\ 1\ \sec$  
(ii)                                                     $when\ t=\frac{1}{3}$  
at rest  $\frac{ds}{dt}=0$                                $s=\left(\frac{1}{3}\right)^3-2\left(\frac{1}{3}\right)^2+\frac{1}{3}-1$  
 $3t^2-4t+1=0$                                $=\frac{-23}{27}=\frac{23}{27}$  
 $\left(t-1\right)\left(3t-1\right)$                           $1-\frac{23}{27}=\frac{4}{27}metres$  
 $t=1\ or\ t=\frac{1}{3}$  

2017 P2 8

(a) The displacement, s , of a particle from a fixed point , O, is given by s =  $t^3-\frac{5}{2}t^2-2t$  metres at time, t seconds. 

(i) Determine the velocity of the particle at t = 3.5 s , clearly stating the correct unit. 
(ii) If the particle is momentarily at rest, find the time, t , at this position. 

(i)  $\frac{ds}{dt}=v$  

 $v=3t^2-5t-2$  
when t = 3.5
 $v=\left(3.5\right)^2-5\left(3.5\right)-2$  
 $=36.75-17.5-2$  
 $=17.25ms^{-1}$  

(ii)
 $\frac{ds}{dt}=0$  
 $3t^2-5t-2=0$  
 $\left(t-2\right)\left(3t+1\right)$  
 $t=2\ or\ t=-\frac{1}{3}$  
t=2 Seconds