1. Feb 19, 2004

### xyxaprilxyx

I have a few questions here... I hope you'll bare with me.

1)A certain car is capable of accelerating at a rate of +.60 m/s^2. How long does it take for his car to go from a speed of 55 mi/h to a speed of 60 mi/h?

can someone please explain to me the step by step way to solve something like this? I dont expect you to solve it for me.. but atleast help me with the formula or equations I need.

2)...if I have an equation such as...
(30m/s - 10m/s)/2.0s = 10m/s^2 (this equation came from the book example)how did the answer get to 10m/s^2 ? I forgot how to.. can someone please enlighten me?

3)...if I have 10m/s^2 and (30m/s -10m/s)only..as in

Last edited: Feb 19, 2004
2. Feb 19, 2004

### HallsofIvy

"1)A certain car is capable of accelerating at a rate of +.60 m/s^2. How long does it take for his car to go from a speed of 55 mi/h to a speed of 60 mi/h?"

The problem with this is that it involves two completely different systems of units. You need to either convert "m/s^2" to "mi/hr^2" or convert mi/h to m/s. Probably the latter is easier. There are, if I remember correctly, 0.62 miles per km so .62 miles per 1000 m or 1000 m per .62 mi= 1000/.62= 1613 meters per mile. That is: 55 mi/h= 55*1613= 88710 m/h and 60 mi/h= 60*1613= 96775 m/hr.
There are 60 minutes per hour and 60 seconds per minute so 3600 seconds per hour. At 55 mi/h, we would be going 88710/3600= 24.6 m/s and, at 60 mi/h, 96775/3600= 26.9 m/s.
Now the problem is "How long does it take to accelerate from 24.6 m/s to 26.9 m/s if the acceleration is 0.60 m/s^2?" That's a change in speed of 26.9- 24.6= 2.3 m/s. Acceleration is "change in speed divided by time" so we are saying 2.3/T= .60 (T is the time in seconds). That is the same as T= 2.3/.6= 4.3 seconds.

"2)...if I have an equation such as...
(30m/s - 10m/s)/2.0s = 10m/s^2 (this equation came from the book example)how did the answer get to 10m/s^2 ? "

Didn't you ask this question in a different forum? There are two things going on here: I take it that the "arithmetic" part-
(30- 10)/2= 20/2= 10 is not the difficulty. It may be keeping track of the units- do that pretty much the way you do numbers: Adding or subtracting the same units keeps them the same (and you CAN'T add or subtract different units!) and multiplying or dividing, use "fraction" rules: m/s divided by s is the same as (m/s)(1/s)= m/s^2.

"3)...if I have 10m/s^2 and (30m/s -10m/s)only..as in
(30m/s - 10m/s)/?= 10m/s^2 how can I solve to find the 2.0s?"

At first I didn't see any difference between this and the previous question! It should be easy to reduce the numerator: 30m/s- 10m/s= 20 m/s. Now the equation is (20m/s)/T= 10 m/s^2. With an equation like "A/x= B", you should think "multiply on both sides by x to get rid of that fraction" so that A= Bx. Now divide by B on both sides so that x is alone: x= A/B. In this problem A= 20 m/s and B= 10 m/s^2
(and x is T of course). A/B= (20 m/s)/(10 m/s^2). 20/10= 2 certainly. (m/s)/(m/s^2) is the same as (fractions: to divide, invert and multiply) (m/s)(s^2/m) or (m/m)(s^2/s)= s:
T= (20m/s)/(10m/s^2)= 2.0 s.

3. Feb 19, 2004

Thanks...