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Calculate the Average speed

  • Thread starter Tonia
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  • #1
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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.
 

Answers and Replies

  • #2
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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?
 
  • #3
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Acceleration = (Vf - V0)/t, Vf = Vo + at
 
  • #4
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How do I find the speed of the car at t=2,5 seconds?
 
  • #5
billy_joule
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With the two equations in post #3.
Find acceleration, then find the velocity after 2.5 seconds of accelerating at that rate.
 
  • #6
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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?
 
  • #7
billy_joule
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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.
 
  • #8
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what do I need to do?
 
  • #9
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I want to understand how to solve this but I'm not sure what I need to do next.
 
  • #10
billy_joule
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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
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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
billy_joule
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No. Reread my post, this part addresses that:
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
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Oh, so it's 45 mi/hr - 0mi/hr divided by 5 seconds? equals 9 mi/hr/sec acceleration?
 
  • #14
billy_joule
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Correct.
 
  • #15
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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
billy_joule
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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
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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
haruspex
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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?
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
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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
haruspex
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9 mi/hr/sec. times 2.5 seconds = 22.5 mi/hr because the seconds cancel??
.
Yes.
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
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So then it's 9 mi/hr/sec. times 2.5 seconds = 22.5 mi/hr??
 
  • #22
haruspex
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So then it's 9 mi/hr/sec. times 2.5 seconds = 22.5 mi/hr??
Yes.
 
  • #23
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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
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what about the time when the speedometer will read the average speed you calculated for part a) as the instantaneous speed??
 
  • #25
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Sorry, I meant that the accleration os 9 mi/hr/sec.
 

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