Math 161

Sample First Midterm

Problem 2

Dr. Wilson

2. A drag racer accelerating from a complete stop, covered 1/4 mile in 5 seconds. A frame by frame analysis of a video tape of the event resulted in the perparation of the following data.

t

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

2.2

2.4

s

0

12.7

29,9

54.9

84.5

118.1

155.2

195.6

238.9

285.1

333.9

385.3

439.0

t

2.6

2.8

3.0

3.2

3.4

3.6

3.8

4.0

4.2

4.4

4.6

4.8

5.0

s

495.0

553.2

613.5

675.8

740.2

806.4

874.6

944.5

1016.2

1089.7

1164.8

1241.6

1320.0

where s denotes the distance traveled in feet and t denotes the time in seconds.

a) Draw the graph. What can you tell about the velocity and acceleration by looking at the graph?

b) The output from a radar gun was included in the video tape. It said that the instantaneous velocity at 4 sec was 241.5 mi/hr. Does this agree with the data above? Compute the average velocity between 3.6 sec. and 4 sec, 3.8 sec. and 4 sec., 4sec. and 4.4 sec., and 4 sec. and 4.2 sec. (Don't forget to change ft./sec. into mi./hr.) What can you say about the acceleration from these results?

c) In order to get a better approximation for the instantaneous velocity, one would need to use smaller increments of time. The video tape has 50 frames per second, so we can get the following refinement of the data.

t

3.80

3.82

3.84

3.86

3.88

3.90

3.92

3.94

3.96

3.98

4.00

s

874.6

881.5

888.4

895.4

902.3

909.3

916.3

923.3

930.4

937.4

944.5

t

4.00

4.02

4.04

4.06

4.08

4.10

4.12

4.14

4.16

4.18

4.20

s

944.5

951.6

958.7

965.8

973.0

980.2

987.3

994.5

1001.7

1009.0

1016.2

What is the best estimate for the instantaneous velocity at 4 sec. that you can get from this data?

d) Notice that in part b) the instantaneous velocity at 4 sec. was fairly close to the average between the average velocity between 3.8 and 4 sec., and between 4 and 4.2 sec. Use this technique to find the instantaneous velocities at 1, 2, and 3 sec. Use these data together with your instantaneous velocity at 4 sec. and the fact that you can figure that the instantaneous velocity when t = 0 is 0, to draw a graph of the velocity as a function of time. What can you say about the acceleration?

a) The graph looks like

We see from the graph that the distance increases with time, which means that the velocity is positive.

We also see that the graph of the distance is curving up, which means that the slope of the distance graph is increasing which means that the velocity is increasing which means that the acceleration is positive. Whether the acceleration is increasing or decreasing is not as evident from the graph of the distance function. It would be better if we could look at the graph of the velocity function to tell whether the acceleration is increasing or decreasing.

b) To find the velocity at a point, we consider the average velocity which obtained from the formula

If we wanted to get the averaage3 velocity between 3 and 4 seconds, we would divide the distance the car traveled in that time by the 1 sedcond difference in time and get

944.5 - 613.5 = 331 ft.

331 ft. in one second is a velocity of 331 ft./sec.

However, if we compute the average velocity between 4 and 5 seconds we get that the car traveled

1320 - 944.5 = 375.5 ft.

for an average velocity of 375.5 ft./sec.between 4 and 5 seconds.

We see that the car was traveling faster between 4 and 5 seconds then it was between 3 and 4 seconds. This is consistent with the fact that the car is traveling faster as it spends more time speeding up. However, the question is what was its instantaneous velocity i.e., exactly how fast was it traveling at 4 seconds. At this point, we can conclude that it is somewhere between 331 ft./sec. and 375.5 ft./sec. The radar gun gave us a speed of 241.5 mi./hr. To see how this compares with our results, we would need to convert mi./hr. to ft./sec. Since there are 5280 ft./mi. and 3600 sec./hr, we multiply the number of miles per hour by

241.5 mi./hr. x 5280 ft./mi. / (3600 sec./hr.) = 354.2 ft./sec.

which is between 331ft./sec. and 375.5 ft./sec. It is not exactly halfway between these two velocities, being just slightly closer to the later velocity than to the earlier velocity, but the average between these two velocities

(331 ft./sec. + 375.5 ft./sec.) / 2 = 353.25 ft./sec.

would give us a fairly close estimate of the instantaneous velocity. However, we would do better if we were to use a smaller time interval. We have enough data to get a better estimate if we use smaller time intervals around 4 seconds. The question asks us to compute the average velocity between 3.6 sec. and 4 sec.

(806.4 ft. - 944.5 ft.)/ (3.6 sec. - 4 sec.) = 345.25 ft./sec.

which is a better estimate of the instantaneous velocity of 354.2 as measured by the radar gun. Again, notice that it is below the instantaneous velocity from the radar gun. If we computed the average velocity between 4 sec. and 4.4 sec. we could get

(1089.7 ft. - 944.5 ft.) / ( 4.4 sec - 4 sec.) = 363 ft. /sec.

which is very close to being as far above the instantaneous velocity as the previous result was below the instantaneous velocity. Again while the instantaneous velocity is a little closer to the higher result, it is much closer to being in the middle of the two which is computed by

(345.25 ft./sec. + 363 ft./sec.) / 2 = 354.125

which is much closer to the instantaneous velocity.

We could do somewhat better if we were to use the intervals between 3.8 sec. and 4 sec. and 4 sec. and 4.2 sec.

(874.6 ft. - 944.5 ft.) / (3.8 sec. - 4 sec.) = 349.5 ft./sec.

and

(944.5 ft. - 1016.2 ft.) / (4 sec - 4.2 sec.) = 358.5 ft./sec.

Again notice that these are both closer to the instantaneous velocity. If we take their averages we get

(349.5 ft./sec. + 358.5 ft./sec.) / 2 = 354 ft./sec.

which is even closer to the instantaneous velocity of 354.2 ft./sec. Again notice that the instantaneous velocity is just above the average of the two which would put it closer to the higher, later velocity than the earlier, lower velocity.

Since the velocity is going up, the acceleration is positive.

c) To get a better approximation of the instantaneous velocity we would need to take smaller intervals of time and distance. We have the refined data from deeper analysis of the video tape results. We could get a better estimate of the instantaneous velocity using the refined data. The best estimate of the instantaneous velocity would be to average the velocities we compute using the interval between 3.98 sec. and 4 sec. and between 4 sec. and 4.02 sec. The first is

(937.4 ft. - 944.5 ft.) / (3.98 sec. - 4 sec.) = 355 ft./sec

and the later one is

( 944.5 ft - 951.6 ft.) / 4 sec. - 4.02 sec.) = 355 ft.sec.

We see that the two agree. They are both above the speed of the radar gun. This is because of round off error. We have gotten down to where we are within round off error of the actual instantaneous velocity.

d) If we use the technique of averaging the velocities from .2 sec. before the time in question and the velocity to ,2 sec. after the time in question to estimate the instantaneous velocity at the time in question, we get the following results for 1, 2, and 3 sec.

(84.5 ft. - 118.1 ft.) / (0.8 sec. - 1 sec.) = 168 ft./sec/

and

(118.1 ft - 155.2 ft) / (1 sec. - 1.2 sec.) = 185.5 ft./sec.

The average of these is

(168 ft./sec. + 185 ft./sec/) / 2 = 176.75 ft./sec.

Around 2 sec. we get

(285.1 ft - 333.9 ft.) / 1.8 sec. - 2 sec.) = 244 ft./sec.

amd

(333.9 ft. - 385.3 ft.) / (2 sec. - 2.2 sec.) = 257 ft./sec.

The average of these is

(244 ft./sec. + 257 ft./sec.) / 2 = 250.5 ft./sec.

Around 3 sec. we get

(553.2 ft. - 613.5 ft.) / (2.8 sec. - 3 sec.) = 301.5 ft./sec.

and

(613.5 ft. - 675.8 ft.) / ( 3 sec. - 3.2 sec.) = 311.5 ft./sec.

The average of these is

(301.5 ft,.sec. + 311.5 ft./sec.) / 2 = 306.5 ft./sec.

We already know that the instantaneous velocity at 4 sec. approximated by this method is 354 ft./sec. This gives us the following dada

t

1

2

3

4

v

176.75

250.50

306.50

354.00

We also know that since the car started from rest, the velocity at t = 0 was 0. We can use this data to prepare the following line graph.

 

The graph is increasing, so the acceleration is positive. However, the velocity is leveling off so the acceleration, while remaining positive is decreasing.

This explains how, when you floor the accelerator in your car, the force from the acceleration will push you back into your seat. However, as you continue to speed up, the acceleration lets up and the force pushing you back into your seat eases up as you speed up.