How Do You Calculate Average Velocity and Acceleration in Different Scenarios?

[rhxoehwhfh]

Q 1. A hockey player gains puck at the centre court. He then skates down the opponent’s goal and scores 2 s later. After scoring, he skates back to guard his own team’s goal post, taking 3 s to run down the court. Calculate the player’s average velocity for the entire time period. (Assume that the hockey court is 5.0 x 101 m long). [3 Marks]

Q 2. Tina walks to her school in 10 min. The displacement for that walk is 250 m [N360E]. What was her average velocity? [2 Marks]

Q 3. Vic was watching a car race on TV. At the instant the flag was lowered to start the race, the picture on TV screen goes out due to surge in the power. When the picture come back on TV, the timer on score board reads 75 s. At this point Vic observes that leading car was on opposite side of the racing track (opposite side to that racing was started). The racing track is oval in shape and 6 Km in length. [6 Marks]
a) Determine leading cars average velocity during the time when TV was without picture? ☺☺
b) What are two possible distances leading car traveled when TV was without picture? ☺
c) Given the record for fastest racing car is 450 Km/hr, which is most likely distance-leading car has traveled when TV was without picture? ☺☺
d) Based on your calculation in (c), calculate leading car average speed when TV was without picture? ☺

Q 4. Veronica’s car velocity increases from +5.0 m/s to +40 m/s in 5 s time interval. What is her average acceleration? [2 marks]

Q5. Jim has a standard gearbox car. One day he parked his car on a hill and went to shopping. After shopping when he started his car, it stalled and started to roll backward down the hill. At this instant time car had the velocity of 4.0 m/s down the hill. Fortunately Jim was able to start his car so car started accelerating back up (up the hill). After accelerating for 3.0 s, the car was traveling uphill at 3.5 m/s. Calculate the car’s acceleration once Jim got it started. (Assume that the car’s acceleration was constant).

 

[rhxoehwhfh]

Here is my answer for my problem, can anyone correct me'"?

1. V= 1/2(V+v) -> 12.5m/s & 16.6m/s -> 14.55 m/s (Average Velocity)
2. 250/ 10*60 = 0.4166m/s or 25m/m
3. a) 6000/2 -> 3000/75 -> 40m/s
b) 3km, 3000m
c) 40m/s -> 144km/hr < 450km/hr; Therefore 450km/hr car is most likely distance-leading car.
d) 40* 60* 60m/hr -> 144000m/hr -> 144km/hr

4. 40-5/5 = 35/5 -> 7m/s

5. a= V-v/t -> a= 3.5-4/3 -> acceleration is -0.166

 

[dynamicsolo]

Here is my answer for my problem, can anyone correct me'"?

1. V= 1/2(V+v) -> 12.5m/s & 16.6m/s -> 14.55 m/s (Average Velocity)

Average velocity is defined as net displacement divided by the time interval. It is average speed that is given by total distance covered divided by the time interval.

In either case, your calculation would not be correct. (For one thing, you don't generally get the average velocity by averaging two different average velocities taken over different intervals.) For this question, recall that the skater started at center court and ended up back at his own goal.

2. 250/ 10*60 = 0.4166m/s or 25m/min

This one is fine.

3. a) 6000 m/2 -> 3000 m
3000 m/75 sec -> 40m/s

I broke this into two statements. Be careful about merging all your calculations into one giant statement: it is not only incorrect mathematically, but it needlessly confuses readers, certain of whom will be grading your assignments and exams...

Otherwise, this answer is OK.

b) 3km = 3000m

This is one possible distance the lead car could have traveled. Is there any other distance it could have covered on the oval track that would put it on the far end when the TV viewer was able to watch it again?

c) 40m/s -> 144km/hr < 450km/hr; Therefore 450km/hr car is most likely distance-leading car.

How fast is 450 km/hr in meters per second? How far could a race car moving that fast travel in 75 seconds?

d) 40* 60* 60m/hr -> 144000m/hr -> 144km/hr

This is connected with part (c). Given that the leading car is seen at the far side of the track and that is has traveled the most likely distance in 75 seconds for a near record holding car, what average speed would it have? (Remember the definition of average speed.)

4. 40-5/5 = 35/5 -> 7m/s

Be careful about your units. The average acceleration is

[ (final velocity) - (initial velocity) ]/ time interval , so this gives

[ (+40 m/sec) - (+5 m/sec) ] / 5 sec = (+35 m/sec)/5 sec = +7 m/(sec^2) ,

which are the proper units for acceleration. Your answer is otherwise correct.

5. a= V-v/t -> a= 3.5-4/3 -> acceleration is -0.166

This question illustrates why I made a point of adding the signs of the velocities in the last answer. How have you account for the directions of the car's velocity? Also remember that you want to subtract the initial velocity from the final velocity. (And don't forget to include the units!)