Discover the Acceleration and Angle of Circular Motion | Car Physics Explained

[googooloo]

physic for today please help me!

Homework Statement

A car starts moving from a stay position. The path of its movement has a circular form.
The radius of the circle that the car is moving on, is constantly 0.5 (m/s^2).
a) what is the pure linear acceleration of the car after 15 seconds?
b) what is the angle between the diagram of pure acceleration and diagram of speed of the car?

P.S: i was not sure about the word "pure linear accelerate", but i think the formula is :
pure linear accelerate = radical of [ (linear a)^2 + (radius a)^2 ]

Homework Equations

The Attempt at a Solution

 

[dreit]

Im not quite sure but i would think that you would need to take into account centripetal force and acceleration

http://en.wikipedia.org/wiki/Uniform_circular_motion#Acceleration

 

[tiny-tim]

welcome to pf!

hi googooloo! welcome to pf!

(try using the X² icon just above the Reply box )

what language was this question written in?

i'll assume it meant …

A car starts moving from a stationary position. The path of its movement has a circular form.
The linear (or tangential) acceleration of the car, is constantly 0.5 (m/s²).
a) what is the linear acceleration of the car after 15 seconds?
b) what is the angle between the direction (or vector) of the total acceleration and the direction (or vector) of the speed of the car?​

your formula is for the total acceleration … the linear (or tangential) acceleration is perpendicular to the https://www.physicsforums.com/library.php?do=view_item&itemid=27", so the total acceleration is √(a_l² + a_r²) …

but i think the question is only asking for the linear (tangential) acceleration

(we really need to see the original question to be sure)

 

[googooloo]

hi googooloo! welcome to pf!

(try using the X² icon just above the Reply box )

what language was this question written in?

i'll assume it meant …

A car starts moving from a stationary position. The path of its movement has a circular form.
The linear (or tangential) acceleration of the car, is constantly 0.5 (m/s²).
a) what is the linear acceleration of the car after 15 seconds?
b) what is the angle between the direction (or vector) of the total acceleration and the direction (or vector) of the speed of the car?​

your formula is for the total acceleration … the linear (or tangential) acceleration is perpendicular to the https://www.physicsforums.com/library.php?do=view_item&itemid=27", so the total acceleration is √(a_l² + a_r²) …

but i think the question is only asking for the linear (tangential) acceleration

(we really need to see the original question to be sure)

Yes that is exactly what you said, and sorry for the language
anyways how can i find √(a_l² + a_r²)?

 

[tiny-tim]

a_r is the https://www.physicsforums.com/library.php?do=view_item&itemid=27" , which you should know about

and a_l is simply dv/dt, the rate of change of the speed (the scalar speed, not the vector velocity)

 

[googooloo]

a_r is the https://www.physicsforums.com/library.php?do=view_item&itemid=27" , which you should know about

and a_l is simply dv/dt, the rate of change of the speed (the scalar speed, not the vector velocity)

well i need to find a_t and a_r...
i have constant accelerate about 0.5 m/s² and the time which it required from me about 15 sec.
SO> 0.5=delta V/ delta t=delta v/15
so delta v is 7.5...v₀ is 0 m/s so V is 7.5...
r=30 m
a_r=V²/r=1.875...
a_t is 0.5 m/s²...
so √(a_l² + a_r²) will be √(3.515+0.25)=1.94m/s²

now the 2nd part i know that V and a_t are in same direction so i need to find the angle between √(a_l² + a_r²) and a_l...
should i use 1.94*cos(teta)=a_t??

 

[Redbelly98]

a_r=V²/r=1.875...
a_t is 0.5 m/s²...
so √(a_l² + a_r²) will be √(3.515+0.25)=1.94m/s²

Yes, that's correct.

now the 2nd part i know that V and a_t are in same direction so i need to find the angle between √(a_l² + a_r²) and a_l...
should i use 1.94*cos(teta)=a_t??

Yes.

 

[googooloo]

Yes, that's correct.

Yes.

Thanks a lot then it is solved