Q2.1 The meaning of derivatives and integrals (K)

The accompanying figure shows the velocity of a particle, moving in 1D, as a function of time. Estimate
  1. the acceleration of the particle at t=0 s;
  2. the acceleration of the particle at t=1 s;
  3. the displacement in the interval shown.
Graphic - No title - derivint
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Hint

You need to make the connections between derivatives and slopes; and between integrals and areas.
 

Solution

Reveal
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Solution

The acceleration is the derivative of the velocity with respect to time, and therefore is equal to the gradient of the velocity–time curve. We can estimate from the figure that at $t=0\,{\rm s}$ this gradient is +2.4 ms-2, while at t=1 s it is -1 ms-2.

The displacement is the integral of the velocity with respect to time:

Δx=vdt
It can be interpreted as the area under the velocity–time curve.

The number of rectangles within this area is Nr21. The area (note its units) of a rectangle is Ar=0.1 ms-1×0.5 s=0.05 m. Thus the displacement is

ΔxNrAr1.05 m