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[h]SL 5.9 Kinematic problems involving displacement, velocity and acceleration

[q] **Kinematics**

This whole section refers to the study of one-dimensional movement: Displacement (position in relation to the origin), velocity (speed, but with a direction (can be negative)), and acceleration.

[q]

We can start by noting that, by definition, velocity is the rate of change of position, and acceleration is the rate of change of velocity. Given that the rate of change is equivalent to the derivative, we get:

\[

v(t) = \frac{ds}{dt}

\]

\[

a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}, \quad \int v(t) dt = s(t) + c, \text{ etc.}

\]

[a]

DISTANCE TRAVELLED:

Distance travelled is closely linked to displacement, so we will be able to integrate velocity to help us find distance travelled. However, distance travelled takes into account all changes in displacement, even negative changes (going backwards). You may remember that distance can be found with the area under a velocity curve, but we use absolute value to deal with backwards movement.

[q]

If doing this manually, you split up any integral between each root, so you can make sure you change any negative areas to positive.

Example 2: A particle has velocity function \(v(t) = 1 – 2t \text{ cm/s}\). Find the distance travelled in the first second of motion:

We need to find any roots to the function first:

\[1 – 2t = 0 \Rightarrow t = 0.5.\]

So we have:

\[

\text{Distance Traveled} = \int_0^{0.5} v(t) dt + \int_{0.5}^{1} |v(t)| dt

\]

And we must split the integral into positive areas and add the absolute value of both results.

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