A train travels along a track and its speed (mph) is given s(t)=29t for the first half hour. Its speed is constant and equal to s(t)=29/2 after the first half hour. t is in hours.

Question

How far (miles) does the train travel in the first hour?

I think I am supposed to combine the areas of two shapes.

Answer 1

Topics tested in this question

-Displacement is the area on the velocity graph

-Time intervals can be treated separately and the results may be combined (problem solving)

Consider the two separate time intervals with constant acceleration.

Interval 1:

s(t) = 29 t, 0 h < t < 0.5 h

The displacement of the train is the area between the curve on the graph and the time axis. This is a line so the area between s=29 t, s=0, and t=0.5 would be the area of a triangle A=Δ1/2 b h. This gives the displacement in the first interval as

Δx₁ = 1/2 (1/2 h)(29/2 mi/h) = 29/8 mi

Interval 2:

s(t) = 29/2, 0.5 h < t < 1 h

This is a constant function. The area would be the area of a rectangle. Similarly

Δx₂ = (29/2 mi/h)(1/2 h) = 29/4 mi

Combining the the displacements gives Δx = 29/8 mi + 29/4 mi = 87/8 mi = 10.875 mi.