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November 28, 2014

November 28, 2014

Posted by **sandhu** on Wednesday, November 24, 2010 at 8:37pm.

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

= (38.8-12.6)/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 8:48pmI 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 10:22pmDistance is pi(R) but displacement is only 2R ,the diameter of the semi-circle

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

- physics -
**sandhu**, Thursday, November 25, 2010 at 6:21amThank you for clarifying

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