Distance travel given constant a

[angelcase]

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.

 

[Karmalo]

(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: $$a_{m}=\frac{\Delta v}{\Delta t}$$ where $$\Delta$$ 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.

 

[Tusike]

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.

 

[angelcase]

Karmalo, what is the third kinematic equation?? I am horrible with physics and reading the book is doing no good...

 

[Karmalo]

$$X=X_{0}+V_{0}t+\frac{1}{2}at^{2}$$
Have you ever used this one?

 

[angelcase]

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)??

 

[Karmalo]

$$X$$ is the final position and $$X_{0}$$ is the initial position so you can say $$X-X_{0}$$ is the distance traveled.

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

Note: $$V_{0}$$ is the initial speed

 

[angelcase]

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.

 

[Karmalo]

So we have this: $$X=X_{0}+V_{0}t+\frac{1}{2}at^{2}$$

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

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

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

$$t=\pm\sqrt{\frac{2\cdot320}{3,086}}$$

As time cannot be negative: $$t=14,4(s)$$