Posted by sandhu on .
Posted by sandhu on Wednesday, November 24, 2010 at 4:29am.
The firetruck goes around a 180°, 162 m radius circular curve. It enters the curve with a speed of 12.6 m/s and leaves the curve with a speed of 38.8 m/s. Assuming the speed changes at a constant rate, what is the magnitude of the total acceleration of the firetruck just after it has entered the curve?
physics  MathMate, Wednesday, November 24, 2010 at 8:16am
From the length of the curve (L=ðr), calculate the average tangential acceleration, At
= (38.812.6)/L
Find the radial acceleration due to curvature at entry to curve (v=12.6m/s)
Ar=v²/r
=12.6²/r
Since the two accelerations are perpendicular to each other, add Ar and At vectorially.
physics  sandhu, Wednesday, November 24, 2010 at 7:50pm
I don't understand how you get At = (38.8  12.6)/L.Please explain
Thanks.

physics 
Damon,
I differ somewhat
I believe that At = change in tangential speed / time. I think he meant t not L
get the time from average speed (38.8+12.6)/2 = 25.7 m/s
now it went pi R = pi(162) = 509 meters
so the time in the turn was
t = 509/25.7 = 19.8 seconds
so I get At = (38.8 12.6)/19.8
= 1.32 m/s^2 
physics 
sandhya,
Distance is pi(R) but displacement is only 2R ,the diameter of the semicircle

physics 
Damon,
It is the scalar, not the vector, distance we are interested in for the magnitude of the acceleration.

physics 
sandhu,
Thank you for clarifying