Acceleration Problem 1: At a sports-car rally, a car starting from rest accelerates uniformly at a rate of 9.0 m/sec^2 over a straight-line distance of 100m. The time to beat in this event is 4.5 seconds. Does the driver do it? If not, what must the minimum acceleration be to do so?
We can start by creating position time, velocity time, and acceleration time graphs along with a motion map. For the position time, we know that the object is moving away from its point of origin at an increasing rate, so a parabolic graph would represent this accurately. For the velocity time, we know the velocity at which the car is moving is increasing at a constant rate, so an upward sloping linear line would represent this the most accurately. Because acceleration is the rate of change of velocity and the velocity time graph is linear, the acceleration time graph will be a constant function. The motion map shows dots that slowly grow farther apart due to the increasing speed.
In order for the car to beat the time of 4.5 seconds with an acceleration of 9 m/s^2, Δx from 0 to 4.5 seconds has to be greater than or equal to 100m. We know that acceleration is the change in velocity over time. Hence, acceleration is the slope of the velocity time graph. Since initial velocity is 0, the equation for the velocity time graph is v=9.0t. We can use this equation to draw a quantitative vt graph.
The displacement is the area under the velocity curve, which is a triangle. Hence, we use the triangle area formula to solve for the area. Since we know the time and velocity, we can calculate the displacement. If it is greater than 100 meters, then the driver becomes the new speed racer.
Area = 1/2 b*h
Δx= 1/2 (4.5sec) ( 40.5 m/sec)
Δx = 91.25 meters
Δx= 1/2 (4.5sec) ( 40.5 m/sec)
Δx = 91.25 meters
The car only travels 91.25 meters in 4.5 seconds, so it does not break the record. In order to break the record, Δx has to be greater than 100 when time = 4.5 seconds.
100m = 1/2 (4.5sec) (x)
200m = 4.5x
x= 44.44 m/sec
200m = 4.5x
x= 44.44 m/sec
We now know initial velocity (0 m/sec) and final velocity (44.44 m/sec) and the change in time. We can use this to calculate the acceleration, which would be (Final Velocity - Initial Velocity)/ (Time) = (44.44 m/sec - 0 m/sec)/ 4.5 sec = 9.88 m/sec^2. An acceleration of 9.88 m/sec^2 is needed to tie/break the record on the track.
Acceleration Problem 2: An experimental rocket car starting from rest reaches a speed of 560 km/h after a straight 400 meter run on a level salt flat. Assuming that acceleration is constant (a) what was the time of the run, and (b) what is the magnitude of the acceleration?
The rocket starts at rest and accelerates constantly until it reaches a speed of 560 km/h. To draw the motion map, we start with points closer together and gradually spread them apart. This shows the increasing velocity due to the increasing lengths of the arrows. To draw the position time graph we start at the origin because the object is not displaced before driving. We start with an unsteep, positive slope to show the slow initial velocity and gradually increase the steepness to show the increase in speed. This is best accomplished by a parabolic graph. To draw the velocity time graph, we also start at the origin because the problem indicates the object starts at rest. Because there is a positive acceleration, and acceleration is the slope of the the velocity graph, we can draw a positive sloping line as the velocity increases. The acceleration time graph would just be constant, but the value is still unknown. We do know, however, that it is positive because the object is moving forward with increasing speed. The resulting graphs are pictured below:
We know the rocket-car reaches 560 km/h at the end of its 400 m run. The units are not consistent with the accepted units of m/sec, so we should convert the 560 km/h to m/sec: 560 km/h * 1 h/ 3600 sec * 1000 m / 1 km = 155.56 m/sec. We can use this value to make a quantitative velocity time graph, pictured below.
We know both the displacement and the time, so we use the concept of displacement being area under the velocity time line to solve for time.
A = 1/2 b * h
Δx = 1/2 v * t
400 m = (1/2) (155.56 m/sec) t
t = 5.143 seconds
Δx = 1/2 v * t
400 m = (1/2) (155.56 m/sec) t
t = 5.143 seconds
b) We can use the concept that acceleration is the change in velocity over time to solve for the acceleration of the rocket-car.
a = Δv/t = (155.6 m/sec - 0 m/sec)/ 5.143 sec = 30.247 m/sec^2