Calculating displacement of an accelerating object
Dragsters can achieve an average acceleration of 26.0 m/s
2 . Suppose a dragster accelerates from rest at this rate for 5.56 s
[link] . How far does it travel in this time?
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.)
Strategy
First, let’s draw a sketch
[link] . We are asked to find displacement, which is
x if we take
to be zero. (Think about
as the starting line of a race. It can be anywhere, but we call it zero and measure all other positions relative to it.) We can use the equation
when we identify
,
, and
t from the statement of the problem.
Sketch of an accelerating dragster.
Solution
First, we need to identify the knowns. Starting from rest means that
,
a is given as 26.0 m/s
2 and
t is given as 5.56 s.
Second, we substitute the known values into the equation to solve for the unknown:
Since the initial position and velocity are both zero, this equation simplifies to
Substituting the identified values of
a and
t gives
Significance
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, our answer is reasonable. This is an impressive displacement to cover in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this. If the dragster were given an initial velocity, this would add another term to the distance equation. If the same acceleration and time are used in the equation, the distance covered would be much greater.
What else can we learn by examining the equation
We can see the following relationships:
Displacement depends on the square of the elapsed time when acceleration is not zero. In
[link] , the dragster covers only one-fourth of the total distance in the first half of the elapsed time.
If acceleration is zero, then initial velocity equals average velocity
, and
Solving for final velocity from distance and acceleration
A fourth useful equation can be obtained from another algebraic manipulation of previous equations. If we solve
for
t , we get
Substituting this and
into
, we get
Calculating final velocity
Calculate the final velocity of the dragster in
[link] without using information about time.
Strategy
The equation
is ideally suited to this task because it relates velocities, acceleration, and displacement, and no time information is required.
Solution
First, we identify the known values. We know that
v0 = 0, since the dragster starts from rest. We also know that
x −
x0 = 402 m (this was the answer in
[link] ). The average acceleration was given by
a = 26.0 m/s
2 .
Second, we substitute the knowns into the equation
and solve for
v :
Thus,
Significance
A velocity of 145 m/s is about 522 km/h, or about 324 mi/h, but even this breakneck speed is short of the record for the quarter mile. Also, note that a square root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration.
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