Posted by sandhu on Wednesday, November 24, 2010 at 8:37pm.
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, Wednesday, November 24, 2010 at 8:48pm
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, Wednesday, November 24, 2010 at 10:22pm
Distance is pi(R) but displacement is only 2R ,the diameter of the semicircle

physics  Damon, Thursday, November 25, 2010 at 2:49am
It is the scalar, not the vector, distance we are interested in for the magnitude of the acceleration.

physics  sandhu, Thursday, November 25, 2010 at 6:21am
Thank you for clarifying
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