Determining Instantaneous Velocity from the Slope at a Point: Jet Car

Calculate the velocity of the jet car at a time of 25 s by finding the slope of the $x$ vs. $t$ graph in the graph below.

Figure 2.50 The slope of an

x

vs.

t

graph is velocity. This is shown at two points. Instantaneous velocity at any point is the slope of the tangent at that point.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is illustrated in Figure 2.50, where Q is the point at $t = 25 s$ .
Solution
1. Find the tangent line to the curve at $t = 25 s$ .
2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s.

3. Plug these endpoints into the equation to solve for the slope, $v$ .

$slope = v_{Q} = \frac{{Δ x}_{Q}}{{Δ t}_{Q}} = \frac{(3120 m - 1300 m)}{(32 s - 19 s)}$
2.96

Thus,

$v_{Q} = \frac{1820 m}{13 s} = 140 m/s.$
2.97
Discussion
This is the value given in this figure’s table for $v$ at $t = 25 s$ . The value of 140 m/s for $v_{Q}$ is plotted in Figure 2.50. The entire graph of $v$ vs. $t$ can be obtained in this fashion.

0 of 3 steps open

How did it go?