A vehicle covers a mile (= 5280 ft) in less than a minute, starting and ending at rest and never exceeding 90 miles/hour. Show that its acceleration or deceleration exceeded 6.6 ft/sec^{2}.

**Solution**

Plot velocity (v in ft/sec ) against time ( t in sec ) .

The graph never gets above the line v = 132 ( ft/sec =90 mph).

If the vehicle's acceleration never exceeds 6.6 ft/sec^{2}, then in particular, its acceleration in the first 20 seconds does not exceed 6.6 ft/sec and hence the velocity curve never gets above the line v = 6.6 t, which cuts the line v = 132 at t = 20.

Similarly, if the vehicle's deceleration never exceeds 6.6 ft/sec^{2}, then its deceleration in the last 20 seconds never exceeds 6.6 ft/sec, so the velocity curve it never gets above the line v = - 6.6 (t - 60), which represents constant deceleration at the maximum 6.6 ft/sec^{2} from t = 40 until the finish at t = 60 when it is stationary. In fact the curve must lie strictly under this line, since the vehicle finishes in less than a minute.

Hence the area under the curve, which represents distance travelled, is less than the area of the quadrilateral bounded by t = 0, v = 6.6t, v = 132 and v = -6.6(t - 60), which is 1/2 20. 132 + 20.132 + 1/2 20.132 = 40.132 = 5280 = 1 mile. But we know that the distance travelled was 1 mile. Hence either it accelerated at more than 6.6 ft/sec^{2} at some point in the first 20 seconds, or it decelerated at more than 6.6 ft/sec^{2} at some point in the last 20 seconds.

*Comment. This solution is much easier than it looks. Just draw the graph!*

© John Scholes

jscholes@kalva.demon.co.uk

25 Aug 2001