Distance travel given constant a

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  • #1
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A hot rod can accelerate from 0 to 60 km/h in 5.4 s.
(a) What is its average acceleration, in m/s2, during this time?


(b) How far will it travel during the 5.4 s, assuming its acceleration is constant?


(c) How much time would it require to go a distance of 0.32 km if its acceleration could be maintained at the value in (a)?


for a, first I did a unit conversion for the speed:

60(1000/3600)= 16.667m/s=v

a=delta v/ delta t

a=16.667/5.4=3.086m/s^2

for b:

d=V^2/2a
d=(16.667)^2/2(3.086)=45.006m

for c:

I converted .32km to 320m.

a=3.086m/s^2 since it is constant...but I don't know what equation to use to determine the time to travel the distance.



I tried the values 19.19 and 10.18, both were wrong.
 

Answers and Replies

  • #2
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(a) and (b) results are correct.

a=delta v/ delta t

a=16.667/5.4=3.086m/s^2
If you put it like that it seems that 16.667 is dv and 5.4 is dt. You can do that because you can also say this: [tex]a_{m}=\frac{\Delta v}{\Delta t}[/tex] where [tex]\Delta[/tex] stands for increment, v for velocity and t for time. You can do this when the acceleration is constant.

For (c) just take the third kinematic equation.
 
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  • #3
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Your formula in (b):
d=v^2 / (2a)

is equivalent to:

d=(a*t)^2 / (2a) = (a * t^2) / 2.
By the way, this I think is more commonly used.
 
  • #4
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Karmalo, what is the third kinematic equation?? I am horrible with physics and reading the book is doing no good...
 
  • #5
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[tex]X=X_{0}+V_{0}t+\frac{1}{2}at^{2}[/tex]
Have you ever used this one?
 
  • #6
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I don't know if I have...is x= to distance? In different books and sites they use different variable to represent distance, whether it be s, d or x.

if that is the case would I rearrange it to solve for t?

2(X-Xsub0)/(Vsub0+a)??
 
  • #7
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[tex]X[/tex] is the final position and [tex]X_{0}[/tex] is the initial position so you can say [tex]X-X_{0}[/tex] is the distance traveled.

You can work out the value of the time (t) from the equation.

Note: [tex]V_{0}[/tex] is the initial speed
 
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  • #8
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I am not sure if I am doing this right.

I plugged in:

320-45-006=16.667t+ 1/2(3.086)t^2
for t I found it to be 3.886 then I added the 5.4sec already passed and got 9.286sec...does that sound correct, or am I not understanding.
 
  • #9
17
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So we have this: [tex] X=X_{0}+V_{0}t+\frac{1}{2}at^{2} [/tex]

We can pass [tex]X_{0}[/tex] to the left and we'd have [tex]X-X_{0}=V_{0}t+\frac{1}{2}at^{2}[/tex]

The distance is 320m so [tex]X-X_{0}=320[/tex], we also know the initial speed is 0 so [tex]V_{0}=0[/tex] and the acceleration, which is 3,086 m/s^2

Now we just have to replace this values and we get this: [tex]320=0+\frac{1}{2}\cdot3,086t^{2}[/tex]

[tex]t=\pm\sqrt{\frac{2\cdot320}{3,086}}[/tex]

As time cannot be negative: [tex]t=14,4(s)[/tex]
 

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