Acceleration on a curve [kinematics]

AI Thread Summary
A car initially traveling at 60 km/h changes direction and speed to 35 km/h over 8 seconds, prompting a calculation of average acceleration. The average acceleration is determined to be -3.125 m/s², indicating a deceleration despite the change in direction. Participants emphasize the importance of treating velocity as a vector, suggesting that resolving velocities into components can clarify the change in velocity. Additionally, discussions touch on the dynamics of a car tilting during a turn, questioning how centripetal force interacts with the forces acting on the vehicle. Overall, the conversation highlights the complexities of kinematics in curved motion and the need for careful vector analysis.
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Homework Statement



a car is moving forward at 60 km/h [E 45° N], executes a turn and is now moving at 35 km/h [N 80° W]. If the turn takes 8 seconds to complete the turn what is the average acceleration of the car in m/s?

Homework Equations



v2 = v1 + a(t)

The Attempt at a Solution



v2 = 35
v1 = 60
t = 8
a = ?

v2 = v1+ a(t)

35km/h = 9.72 m/s
60km/h = 16.6 m/s

35 = 60 + a(8)
35-60 = 8a
a= -3.125m/s

i'm not sure if its right, because don't you accelerate positively when you change direction
 
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boogaaaaa said:

Homework Statement



a car is moving forward at 60 km/h [E 45° N], executes a turn and is now moving at 35 km/h [N 80° W]. If the turn takes 8 seconds to complete the turn what is the average acceleration of the car in m/s?

Homework Equations



v2 = v1 + a(t)

The Attempt at a Solution



v2 = 35
v1 = 60
t = 8
a = ?

v2 = v1+ a(t)

35km/h = 9.72 m/s
60km/h = 16.6 m/s

35 = 60 + a(8)
35-60 = 8a
a= -3.125m/s

i'm not sure if its right, because don't you accelerate positively when you change direction

Welcome to PF.

Remember velocity is a vector. As is acceleration. So ... that means that you might do better to resolve the velocities into their components and then figure the change in velocity by x,y to determine the direction and angle of the average change over the 8 seconds.
 
ok well based on what you said: split into x and y and find average i came up with this diagram:

http://img132.imageshack.us/img132/3949/diagram.jpg

I used the previous values (60), (35) to create the x and y for the purple triangle and finally finding the average speed represented by the "?", i then divided that value by the time (8 seconds) and got my answer. Was my method correct?
 
Last edited by a moderator:
boogaaaaa said:
ok well based on what you said: split into x and y and find average i came up with this diagram:

http://img132.imageshack.us/img132/3949/diagram.jpg

I used the previous values (60), (35) to create the x and y for the purple triangle and finally finding the average speed represented by the "?", i then divided that value by the time (8 seconds) and got my answer. Was my method correct?

I prefer to do it algebraically.

Vi = Vix + Viy

Vi = |Vi|*Cosθ x + |Vi|*Sinθ y
 
Last edited by a moderator:
As to your drawing ...

You want to keep in mind that

a = ΔV/Δt

where

ΔV = Vf - Vi
 
alright thank you, solved :D
 
Hi guys,

I have similar problem with boogaaaa.

As we know, if a car accelerate on curve at certain speed, it will tilt outside the circle track. So i sketch this 2 picture to show how it works.

http://img193.imageshack.us/img193/2323/66057246.jpg
http://img194.imageshack.us/img194/9272/96495741.jpg

I just wondering, how can the car tilt or flip outside the circle track if the side force (centripetal force) is pointing inside the circle?

My suggestion:
Total Moment about tyre (2); ∑M2=0

So, Fs*h + G*b/2 - FN1*b = 0

I assume that FN1=0 since the car is about to tilt. So here come the problem.

Fs = - G*b / 2h.

The minus sign shows that the side force, Fs, should point outside the circle.

can anyone help me?
 
Last edited by a moderator:

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