Motion in a straight line with constant acceleration

Recall that acceleration at a given time measures the instantaneous change in velocity – see Quantities & Units in Mechanics. For motion in a straight line with constant acceleration, the speed along the line changes by the same amount every second. Hence, if speed is measured in metres per second (m/s) then acceleration is measured in metres per second per second (m/s/s). For example, consider a ball rolling along a flat surface:
motion in a straight line
If the ball rolls with speed 4 m/s and accelerates at a constant rate of 2 m/s/s, the ball will have speed 10 m/s after 3 seconds. It is important to note, however, that motion in a straight line may not have constant acceleration.


Distance-Time and Speed-Time Graphs

The gradient on a distance-time graph gives you the instantaneous measure of speed.
Similarly, the gradient on a speed-time graph gives you the instantaneous measure of acceleration. The total area beneath the speed-time graph gives you the total distance measured. (TBC)


Calculus of Position, Velocity and Acceleration

Velocity can be thought of as rate of change in position. Similarly, acceleration can be thought as rate of change in velocity. If position is given as a function of time, say r(t), velocity can be found be differentiating r(t). As a result, velocity is generally also a function of time. Analogously, acceleration a(t) can be found by differentiating velocity v(t). It follows that acceleration can be found by differentiating r(t) with respect to t twice. Mathematically, this can be written as

v(t)=\frac{dr(t)}{dt},\hspace{20pt}a(t)=\frac{dv(t)}{dt}=\frac{d^2r(t)}{dt^2}

Conversely, if velocity v(t) is given in terms of t,  integrating v(t) will give an expression for r(t). Similarly, if acceleration a(t) is given in terms of t, integrating a(t) will give v(t). It follows that r(t) can be found by integrating a(t) twice. Mathematically:

r(t)=\int v(t)\, dt=\int\left(\int a(t)\, dt\right)dt,\hspace{10pt}v(t)=\int a(t)\, dt

Note that when integrating, various integration constants may appear. Evidently, it is essential to use the information from the question to find the values of these integration constants.

Example 1

Recall that the SUVAT equation that relates displacement (S), initial velocity (U), time (T) and constant acceleration (A) is given by

S=UT+\frac{1}{2}AT^2

See SUVAT equations. For a particle with a constant acceleration of 4.7 m/s/s and an initial velocity of 7.05 m/s, use this equation to show that the final velocity of the particle is given by

V=\frac{47}{20}(2T+3)


Example 2

A boomerang is thrown and is modelled as a particle moving along a straight line. The acceleration of the particle is modelled by the equation

a(t)=-2t+2

Given that the boomerang is thrown with an initial speed of 8 m/s, find the position of the boomerang when it turns back on itself.