Calculating Velocity and Using Vector Diagrams in Curved Motion

[bobsagget]

A car is traveling 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?

 

[tiny-tim]

Welcome to PF!

Hi bobsagget ! Welcome to PF!

(try using the X² and X₂ tags just above the Reply box )

A car is traveling 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.
…
so i thought that u can solve this by just plugging into the formula

a= v2-(-v1)/ delta T

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. ​

 

[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

 

[tiny-tim]

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

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

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

 

[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?

 

[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?

 

[tiny-tim]

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?

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?

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? ​

 

[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

 

[tiny-tim]

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

You need to be more accurate about the direction …

it's not exactly NW …

what angle west of north is it?

 

[bobsagget]

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

 

[tiny-tim]

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

(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. ​