# Homework Help: Acceleration question HELP?

1. Oct 13, 2009

### bobsagget

A car is travelling at 19.0 m/s [W] when it enters a curved portion of the track and experiences an average acceleration of 0.270m/s^2 [N] for 62.0 seconds. Determine the velocity of the car after this acceleration. Include a vector diagram.

-v1= 19.0m/s [E]
a= 0.270 m/s^2 [N]
delta t= 62.0 s
v2= ?

so i thought that u can solve this by just plugging into the formula

a= v2-(-v1)/ delta T
which gives me 36 m/s as v2, but it also says to use a vector diagram, which i get another answer of 25.3 m/s? HELP?

2. Oct 13, 2009

### tiny-tim

Welcome to PF!

Hi bobsagget ! Welcome to PF!

(try using the X2 and X2 tags just above the Reply box )
Yes, you can, but remember velocities are vectors, and so obey the law of vector addition (and subtraction) …

you can't just add the magnitudes.

3. Oct 13, 2009

### bobsagget

hey thanks for the info

so do you have to swap the v1 to a -v1 so u can add the v2 and v1 together? like switch the the heading it has to its opposite

4. Oct 14, 2009

### tiny-tim

No, v1 is west and (v2 - v1) is north,

so to do it the vector way, either add components, or use a vector triangle

5. Oct 14, 2009

### bobsagget

so if i were to put the v1 and deltav(v2-v1) into a vector triangle it would be like

HYP= v2 lDelta v
l
________ l
v1

then to solve its v2= delta v^2-v1^2
v2=(16.7)^2 -(19)^2
which would give me 25.3? is that correct?

6. Oct 14, 2009

### bobsagget

tried to make a traingle, but didnt work, but delta v is noorth and your v1 is west, and u have to solve the hypotenuse which is v2?

7. Oct 14, 2009

### tiny-tim

Yes, 25.3 is correct (and you meant +, not - )

How did your triangle not work?

the original velocity, 19, is west. The "added" velocity, 16.7, is north, and so the final velocity is the hypotenuse.

Why is that worrying you?

8. Oct 14, 2009

### bobsagget

yes it was a + your right, I just made a mistake with the whole diagram thing but I got it thanks for your help, and then the heading of this is NW

9. Oct 14, 2009

### tiny-tim

You need to be more accurate about the direction …

it's not exactly NW …

what angle west of north is it?

10. Oct 14, 2009

### bobsagget

off the north line I got the angle to be 49 degrees, or w41n

11. Oct 15, 2009

### tiny-tim

(i've never seen "w41n", though i have seen "n49w", but if your professor says it's ok then of course it's ok)

Yup!

So the full answer is 25.3 m/s 49º West of North.