What is the Average Speed and Acceleration of a Car on a Straight Road?

AI Thread Summary
The average speed of a car accelerating from rest to 45 miles per hour over a 5-second interval is calculated to be 22.5 miles per hour. The constant acceleration during this period is determined to be 9 miles per hour per second. At 2.5 seconds, the speed of the car is also 22.5 miles per hour, which matches the average speed due to uniform acceleration. The time when the speedometer reads the average speed as instantaneous speed is at 2.5 seconds. Overall, the calculations confirm that average speed, instantaneous speed at half the time, and acceleration are interrelated in this scenario.
Tonia
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Homework Statement


On a straight road you start your car and increase the speed at a steady rate to 45 miles per hour during a 5 second interval. Calculate:
a)the average speed during the 5 seconds
b) the acceleration in miles per second at t = 3.2 sec. after you start your car
c) the speed of the car at t = 2.5 sec.
d) the time when the speedometer will read the average speed you calculated for part a) as the instantaneous speed.

Homework Equations


Ave. speed = Vf + Vo/2 = (45 mi/hr + 0 mi/hr)/2 = (45mi/hr)/2 = 22.5 mi/hr

The Attempt at a Solution


Ave. speed = Vf + Vo/2 = (45 mi/hr + 0 mi/hr)/2 = (45mi/hr)/2 = 22.5 mi/hr
The speed of the car at t = 2.5 seconds is: change in distance divided by change in time
I am not sure how to find the acceleration, or the time when the speedomoter will read the average speed in part a) as the instantaneous speed.
 
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Acceleration should be in miles per second per second (mi/s##^2##), perhaps you mistyped. In the question it says that the speed is increased at a steady rate, so the acceleration should be the same at any time t<5, so you should be able to work it out from the initial and final speeds.

For the time when the speedometer reads the average speed, imagine this as your final speed since it is the number you're interested in. So you have an initial speed, a final speed and an acceleration. Do you have an equation that relates these quantities to time?
 
Acceleration = (Vf - V0)/t, Vf = Vo + at
 
How do I find the speed of the car at t=2,5 seconds?
 
With the two equations in post #3.
Find acceleration, then find the velocity after 2.5 seconds of accelerating at that rate.
 
the acceleration in mil/sec. = (Vf - Vo)/t = 45 mi/hr divided by 3.2 sec. = 14.1 mi/hr.
Velocity after 2.5 seconds of accelerating at that rate = Vf = Vo +at = 0 + 14.1 mi/hr times 2.5 = 35.25 mi/sec.
Now what about the speed of the car at 2.5 seconds?
 
Tonia said:
the acceleration in mil/sec. = (Vf - Vo)/t = 45 mi/hr divided by 3.2 sec. = 14.1 mi/hr.

That does not follow the problem statement:
"On a straight road you start your car and increase the speed at a steady rate to 45 miles per hour during a 5 second interval."
And your units don't make sense.

Now what about the speed of the car at 2.5 seconds?
In this case it is the same as the velocity.
 
what do I need to do?
 
I want to understand how to solve this but I'm not sure what I need to do next.
 
  • #10
Tonia said:
the acceleration in mil/sec. = (Vf - Vo)/t = 45 mi/hr divided by 3.2 sec. = 14.1 mi/hr.

You have correctly identified Vf and Vo (45mi/hr and 0 mi/hr)
the time taken to go from Vo to Vf is not 3.2 seconds, The correct time value appears in the first line of the problem statement...

The problem with the units is the seconds disappeared from your answer:

""(Vf - Vo)/t = 45 mi/hr divided by 3.2 sec. = 14.1 mi/hr.""

You wanted to find an acceleration, but mi/hr is a velocity...

if you divide mi/hr by seconds you get mi/hr/s which is an acceleration - a change in velocity (mi/hr) per time (seconds).

Qb asks for an acceleration with units of velocity...there must be a typo somewhere, re-check the question.
( it's also a bit of a trick question, as is, because the answer is the same for all times between 0 and 5 seconds which includes, of course, 3.2 seconds. we know this because the problem statement says "increase the speed at a steady rate" )
 
  • #11
So I just do this: (Vf - Vo)/t = 45 mi/hr divided by 3.2 seconds and I get 14.1 mi/hr/sec. which is the acceleration?
 
  • #12
No. Reread my post, this part addresses that:
billy_joule said:
You have correctly identified Vf and Vo (45mi/hr and 0 mi/hr)
the time taken to go from Vo to Vf is not 3.2 seconds, the correct time value appears in the first line of the problem statement...
 
  • #13
Oh, so it's 45 mi/hr - 0mi/hr divided by 5 seconds? equals 9 mi/hr/sec acceleration?
 
  • #14
Correct.
 
  • #15
a) the average speed during the 5 seconds
b) the acceleration in miles per second at t = 3.2 sec. after you start your car = 9 mi/hr/sec??
c) the speed of the car at t = 2.5 sec.
d) the time when the speedometer will read the average speed you calculated for part a) as the instantaneous speed.
 
  • #16
A and b are correct, assuming they meant miles per hour per second for b. C & d can both be solved using this equation:
Velocity = acceleration *time
 
  • #17
So, for c): the speed of the car at t = 2.5 seconds = velocity = accel. times the time = 9 mi/hr/sec. times 2.5 seconds = 22.5 mi/hr/sec^2 and for d) it's the same answer??
How does speed equal the velocity which equals the acceleration times the time?? I don't get that.
 
  • #18
Tonia said:
9 mi/hr/sec. times 2.5 seconds = 22.5 mi/hr/sec^2
mi/hr/sec times sec is not going to yield mi/hr/sec^2. What should the units be?
Tonia said:
How does speed equal the velocity which equals the acceleration times the time?? I don't get that.
Velocity is a vector, so has a direction. Speed is magnitude of velocity. So they cannot be said to be equal.
In the present problem, all happens in one dimension, so you might as well work on terms of speed.
But you need to distinguish average speed from instantaneous speed. If the acceleration is constant, a, then the instantaneous speed after time t, starting from rest, is at. But the average speed to time t is half that.
 
  • #19
9 mi/hr/sec. times 2.5 seconds = 22.5 mi/hr because the seconds cancel??
So I get the instantaneous from multiplying the acceleration times the time? which would be 9 mi/hr/sec. times 5 seconds = 45 mi/hr?? I am confused.
 
  • #20
Tonia said:
9 mi/hr/sec. times 2.5 seconds = 22.5 mi/hr because the seconds cancel??
.
Yes.
Tonia said:
So I get the instantaneous from multiplying the acceleration times the time? which would be 9 mi/hr/sec. times 5 seconds = 45 mi/hr?? I am confused.
Yes, that's how you would find the instantaneous speed at 5 seconds, but for part c) you want the instantaneous speed at 2.5 seconds.
 
  • #21
So then it's 9 mi/hr/sec. times 2.5 seconds = 22.5 mi/hr??
 
  • #22
Tonia said:
So then it's 9 mi/hr/sec. times 2.5 seconds = 22.5 mi/hr??
Yes.
 
  • #23
So then, the average speed is the same as the acceleration and the same as the instantaneous speed? which is all 22.5 mi/hr??
 
  • #24
what about the time when the speedometer will read the average speed you calculated for part a) as the instantaneous speed??
 
  • #25
Sorry, I meant that the accleration os 9 mi/hr/sec.
 
  • #26
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??
 
  • #27
Tonia said:
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.
Tonia said:
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?
 
  • #28
The graph would look like this: /
 
  • #29
Would the expression for speed be: v = m = (45 mi/hr - 0 mi/hr)/2.5 sec. = 18 mi/hr/sec.??
 
  • #30
Tonia said:
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)?
 
  • #31
It would be the same?
 
  • #32
Tonia said:
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.
 
  • #33
Tonia said:
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.
(Please use the Reply button or the quote button so people know what you are answering.)
 
  • #34
The equation would be Vf = Vo + at??
 
  • #35
Tonia said:
The equation would be Vf = Vo + at??
Right, so plug in the values for v0, vf and a that are appropriate to question d).
 
  • #36
45 mi/hr = 0 + 9 mi/hr/sec times t
Solve for t??
 
  • #37
t would be 5.
 
  • #38
Tonia said:
45 mi/hr = 0 + 9 mi/hr/sec times t
Solve for t??
What was the speed you calculated in part a)?
 
  • #39
22.5 mi/hr
 
  • #40
Tonia said:
22.5 mi/hr
Right. That is the speed you are told to use in answering part d).
 
  • #41
so t is 2.5 seconds and that's the answer to d.
 
  • #42
Tonia said:
so t is 2.5 seconds and that's the answer to d.
Yes.
 
  • #43
Okay. thanks for your help!
 
  • #44
Let me ask you one more question just to make sure: for problems a) and c), the answer is the same? 22.5mi/hr??
 
  • #45
Tonia said:
Let me ask you one more question just to make sure: for problems a) and c), the answer is the same? 22.5mi/hr??
Yes.
 
  • #46
Okay, thanks!
 

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