Acceleration

Example 2.4 Calculating Acceleration: A Subway Train Speeding Up

Suppose the train in Figure 2.18(a) accelerates from rest to 30.0 km/h in the first 20.0 s of its motion. What is its average acceleration during that time interval?

Strategy

It is worth it at this point to make a simple sketch:

A point represents the initial velocity of 0 kilometers per second. Below the point is a velocity vector arrow pointing to the right, representing the final velocity of thirty point zero kilometers per hour. Below the velocity vector is an acceleration vector arrow labeled a equals question mark.

Figure 2.19

This problem involves three steps. First we must determine the change in velocity, then we must determine the change in time, and finally we use these values to calculate the acceleration.

Solution

Identify the knowns. \(v_{0}=0\) (the trains starts at rest), \(v_{\mathrm{f}}=30.0 \mathrm{~km} / \mathrm{h}\), and \(\Delta t=20.0 \mathrm{~s}\).
Calculate \(\Delta v\). Since the train starts from rest, its change in velocity is \(\Delta v=+30.0 \mathrm{~km} / \mathrm{h}\), where the plus sign means velocity to the right.
Plug in known values and solve for the unknown, \(\bar{a}\).

\(\bar{a}=\frac{\Delta v}{\Delta t}=\frac{+30.0 \mathrm{~km} / \mathrm{h}}{20.0 \mathrm{~s}}\)
Since the units are mixed (we have both hours and seconds for time), we need to convert everything into SI units of meters and seconds. (See Physical Quantities and Units for more guidance).
\(\bar{a}=\left(\frac{+30 \mathrm{~km} / \mathrm{h}}{20.0 \mathrm{~s}}\right)\left(\frac{10^{3} \mathrm{~m}}{1 \mathrm{~km}}\right)\left(\frac{1 \mathrm{~h}}{3600 \mathrm{~s}}\right)=0.417 \mathrm{~m} / \mathrm{s}^{2}\)

Discussion

The plus sign means that acceleration is to the right. This is reasonable because the train starts from rest and ends up with a velocity to the right (also positive). So acceleration is in the same direction as the change in velocity, as is always the case.