- #26

- 96

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Tonia
- Start date

- #26

- 96

- 0

- #27

- 36,796

- 7,126

The instantaneous speed at half time turned out to be the same as the average speed over the whole time. For uniform acceleration, that is what you would expect.But the instantaneous speed appears to be the same number as the average speed, which is 22.5 mi/hr.

Draw a graph of speed against time. With uniform acceleration from rest, what will the graph look like? Compare the average height of the graph (average speed) with the height in the middle.

If the acceleration is 9mi/hr/sec, can you write an expression for the speed in mi/hr at time t seconds?what about the time when the speedometer will read the average speed you calculated for part a) as the instantaneous speed?

- #28

- 96

- 0

The graph would look like this: /

- #29

- 96

- 0

Would the expression for speed be: v = m = (45 mi/hr - 0 mi/hr)/2.5 sec. = 18 mi/hr/sec.??

- #30

- 36,796

- 7,126

Right, so how would the average height (=average speed) compare with the height half way along (=instantaneous speed at half time)?The graph would look like this: /

- #31

- 96

- 0

It would be the same?

- #32

- 36,796

- 7,126

No, that's calculating an acceleration, not a speed.Would the expression for speed be: v = m = (45 mi/hr - 0 mi/hr)/2.5 sec. = 18 mi/hr/sec.??

Write out the general equation that relates speed, time and (uniform) acceleration.

You have a value for the acceleration, and a value for the speed, and we're leaving the time as an unknown t. Plug those into the equation.

- #33

- 36,796

- 7,126

What would be the same as what?It would be the same?

Edit: Ok, you were replying to this:

Right, so how would the average height (=average speed) compare with the height half way along (=instantaneous speed at half time)?

Yes.

(Please use the Reply button or the quote button so people know what you are answering.)

- #34

- 96

- 0

The equation would be Vf = Vo + at??

- #35

- 36,796

- 7,126

Right, so plug in the values for vThe equation would be Vf = Vo + at??

- #36

- 96

- 0

45 mi/hr = 0 + 9 mi/hr/sec times t

Solve for t??

Solve for t??

- #37

- 96

- 0

t would be 5.

- #38

- 36,796

- 7,126

What was the speed you calculated in part a)?45 mi/hr = 0 + 9 mi/hr/sec times t

Solve for t??

- #39

- 96

- 0

22.5 mi/hr

- #40

- 36,796

- 7,126

Right. That is the speed you are told to use in answering part d).22.5 mi/hr

- #41

- 96

- 0

so t is 2.5 seconds and that's the answer to d.

- #42

- 36,796

- 7,126

Yes.so t is 2.5 seconds and that's the answer to d.

- #43

- 96

- 0

Okay. thanks for your help!

- #44

- 96

- 0

- #45

- 36,796

- 7,126

Yes.

- #46

- 96

- 0

Okay, thanks!

Share: