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Speed/Velocity Calculations

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Speed/Velocity Calculations Answer Key

Instructions: Read each question carefully. Determine what the question is asking and the steps you need to take to solve the problem. For each calculation, you must show all your math work.

1. 
A kid drives 4 miles to the mall. If the speed limit is 45 miles/hr and the kid makes the trip in .03 hours, is the kid breaking the speed limit?
  1. No, he is going 1 mile/hr.
  2. Yes, he is going 133 miles/hr.
  3. No, he is going 40 miles/hr.
  4. Yes, he is going 150 miles/hr.
2. 
If a car travels at an average speed of 50kms/hr, how much time would it take to complete a trip of 75 kilometers?
  1. 30 mins
  2. 50 mins
  3. 60 mins
  4. 90 mins
3. 
Two bicyclists approach each other on the same road. One has a speed of 3 m/s, and the other has a speed of 9 m/s. They are 530 m apart. If they keep traveling at the same speeds, how long will it be before they meet each other?

[math]"Given:"[/math]
[math]d_1+d_2=530m[/math]

[math]v_1=(d_1)/(t_1)[/math]

[math]v_2=(d_2)/(t_2)[/math]

[math]t_1=t_2=t[/math]
  • [math]d_1=v_1t " and "d_2=v_2t[/math]

    Substitute: [math]v_1t+v_2t=530m[/math]

    [math]t(v_1+v_2)=530m[/math]

    [math]t=(530m)/(v_1+v_2)=(530m)/(3m/s+9m/s)=44.2s[/math]
4. 
Khalil is driving laps at a racetrack. The track is circular and has a radius of 25 m. After completing 2 full laps, what is Khalil’s displacement relative to his starting position? What is total distance traveled in the two laps and how do you determine it? Explain the difference in your answers.
  • The total displacement is the difference in starting position vs. ending position. Since he ended where the race started, his total displacement is equal to zero. Using the formula for circumference of a circle, [math]C=2pir[/math], you can find the length of one track lap and then multiply by 2 to find Khalil"s total distance traveled.

    [math]C=2*pi*25m=157.1m[/math] and [math]157.1m*2=314.2m[/math], which is the total distance traveled by Khalil.
5. 
At a family reunion, the Jackson side of the family is competing with the Williams side of the family in a game of tug-of-war. The Jackson side of the family first loses ground, until x = −12 m, then gains it until x = 42 m, when the Williams side of the family loses its grip. In order to win, the total displacement of team must be 50 m or more. Did the Jackson side of the family win the round?
  • The Jackson side of the family's displacement is [math]x_f-x_i=42m[/math]. Displacement does not depend on the path taken, it is just the difference between final position and initial position. Thus the Jackson side of the family was displaced only 42 m, not enough to win this competition.

    **Potential error by students would be to use the -12m in their calculation of displacement instead of the initial point of 0.
6. 
An object that starts at a position of 13 m and travels for 6 s at a velocity of −9 m/s ends up at what position?
  • [math]d=vt=-9m/s*3s=-54m[/math]

    [math]-54m+13m=-41m[/math]
7. 
The ostrich, the fastest two-legged land animal, has been clocked at a speed of 19.4m/s when being pursued by a predator. An ostrich got loose from a ranch in west Texas and is being chased by a coyote (also able to run just as fast) through a 35mph school zone being clocked by the local sheriff. Could the ostrich and coyote get speeding tickets? Assume it is on a school day during school zone hours.
  • Relationships: 1 mile = 1,609 m

    [math](19.4m/s)*(1" mile")/(1609m)*(60" seconds")/(1" minute")*(60" minutes")/(1" hour")=43.5" mph"[/math]

    So, yes, both the ostrich and the coyote could be issued speeding tickets because [math]43.5" mph">35" mph"[/math].
8. 
Pedro drove 67 km due west and then 94 km due east. The trip took him 2.75 hours to complete. What is his total displacement, total distance, and average speed during the drive?
  • Displacement = [math]Sigma[/math] displacements
    [math]-67km" west "+94km" east "=27km" east"[/math]

    Distance = [math]Sigma[/math] the individual distances
    [math]67km+94km=161km[/math]

    Average speed = distance/time

    [math]s=(161km)/(2.75hr)=58.5(km)/(hr)[/math]
9. 
At the point when the race clock says 00:05:15.00 (5min 15sec), one car is 500 m behind the starting line. By the time the clock reaches 00:05:56:00, the car is 800 m in front of the starting line. What is the car’s average velocity during this time in mph? Assume the car moves in a straight line at constant speed in the positive direction. (1m=0.001km and 1km/hr=0.621371mph)
  • [math]v=x/t[/math]

    [math]x=x_f-x_i=800m-(-500m)=1300m[/math]

    [math]t=t_f-t_i=56s-15s=41s[/math]

    [math]v=(1300m)/(41s)=31.7m/s[/math]

    [math]31.7m/s*(60s)/(1min)*(60min)/(1hr)*(1km)/(1000m)=114.12(km)/(hr)[/math]

    Conversion: [math]1(km)/(hr)=0.621371mph[/math]

    [math](114.12)*(0.621371mph)=70.91mph[/math]
10. 
Devin is driving 35 m/s and after seeing a police officer slows down to 20 m/s in about 5 seconds. He continues at a constant speed for 6 seconds.
-(a) Draw a velocity-time graph of his motion.
-(b) What was his acceleration after seeing the police officer? (include sign and units)
Graph 20x20 Quadrant 1
  • -3 m/s/s

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