Calculating Displacement of an Accelerating Object: Dragsters · Physics Playground

Example 2.10 from College Physics, 2.5 Motion Equations for Constant Acceleration in One Dimension

Calculating Displacement of an Accelerating Object: Dragsters

Dragsters can achieve average accelerations of $26 . {0 m/s}^{2}$ . Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time?

Dragster accelerating down a race track.

Figure 2.31 U.S. Army Top Fuel pilot Tony “The Sarge” Schumacher begins a race with a controlled burnout. (credit: Lt. Col. William Thurmond. Photo Courtesy of U.S. Army.)

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
Draw a sketch.

Figure 2.32
We are asked to find displacement, which is $x$ if we take $x_{0}$ to be zero. (Think about it like the starting line of a race. It can be anywhere, but we call it 0 and measure all other positions relative to it.) We can use the equation $x = x_{0} + v_{0} t + \frac{1}{2} {at}^{2}$ once we identify $v_{0}$ , $a$ , and $t$ from the statement of the problem.
Solution
1. Identify the knowns. Starting from rest means that $v_{0} = 0$ , $a$ is given as $26.0$ and $t$ is given as 5.56 s.
2. Plug the known values into the equation to solve for the unknown $x$ :

$x = x_{0} + v_{0} t + \frac{1}{2} {at}^{2} .$
2.41

Since the initial position and velocity are both zero, this simplifies to

$x = \frac{1}{2} {at}^{2} .$
2.42

Substituting the identified values of $a$ and $t$ gives

$x = \frac{1}{2} (26 . {0 m/s}^{2}) {(5 . 56 s)}^{2},$
2.43

yielding

$x = 402 m.$
2.44
Discussion
If we convert 402 m to miles, we find that the distance covered is very close to one quarter of a mile, the standard distance for drag racing. So the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this.

0 of 3 steps open

How did it go?

Strategy

Solution

Discussion