Rotational bodies

jddery · #1 $Old$ November 12th, 2009, 02:24 PM

A car turns a corner with a radius of curvature of 18.2 m while braking to reduce its speed. If the brakes generate an angular deceleration of 0.7 rad/s2 what is the magnitude of the acceleration of the car half way through the corner when the car's linear speed is 4.2 m/s?

[Specify units of 'm/s2' in your answer.]

skeeter · #2 $Old$ November 12th, 2009, 04:45 PM

Quote:

Originally Posted by jddery $View Post$

A car turns a corner with a radius of curvature of 18.2 m while braking to reduce its speed. If the brakes generate an angular deceleration of 0.7 rad/s2 what is the magnitude of the acceleration of the car half way through the corner when the car's linear speed is 4.2 m/s?

[Specify units of 'm/s2' in your answer.]

$|a| = \sqrt{a_T^2 + a_c^2}$

where tangential acceleration is $a_T = r\alpha$

and centripetal acceleration is $a_c = \frac{v^2}{r}$

jddery · #3 $Old$ November 12th, 2009, 06:23 PM

hey thanks that worked

now how did you come up with that equation? is that a defined equation in a text book?
or did you simply know that adding the tangential acceleration with the centripetal acceleration by means of Pythagorus would result in acceleration?

mathaddict · #4 $Old$ November 12th, 2009, 09:18 PM

Quote:

Originally Posted by jddery $View Post$

hey thanks that worked

now how did you come up with that equation? is that a defined equation in a text book?
or did you simply know that adding the tangential acceleration with the centripetal acceleration by means of Pythagorus would result in acceleration?

THe car experiences 2 types of acceleration ie tangential a and centripetal a (towards the centre) . So its resultant acceleration would be the vector sum of the 2 accelerations . Phythagoras would give you the result since the 2 accelrations are 90 to each other .