# Calculate the Average speed

But the instantaneous speed appears to be the same number as the average speed, which is 22.5 mi/hr. Now what about: what about the time when the speedometer will read the average speed you calculated for part a) as the instantaneous speed??

haruspex
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But the instantaneous speed appears to be the same number as the average speed, which is 22.5 mi/hr.
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.
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.
what about the time when the speedometer will read the average speed you calculated for part a) as the instantaneous speed?
If the acceleration is 9mi/hr/sec, can you write an expression for the speed in mi/hr at time t seconds?

The graph would look like this: /

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

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

It would be the same?

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

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

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.

The equation would be Vf = Vo + at??

haruspex
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The equation would be Vf = Vo + at??
Right, so plug in the values for v0, vf and a that are appropriate to question d).

45 mi/hr = 0 + 9 mi/hr/sec times t
Solve for t??

t would be 5.

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

22.5 mi/hr

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

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

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

Let me ask you one more question just to make sure: for problems a) and c), the answer is the same? 22.5mi/hr??

haruspex