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DISCUSSION and HINTS
As expected, the acceleration of P has both non-zero tangential and normal components.
- From the equation provided for speed as a function of distance traveled, the speed of P is monotonically decreasing over the range of motion shown in the animation below. Therefore, the tangential component of acceleration always points "backward" of the direction of motion.
- The normal component decreases as P moves along the track since the speed of P is decreasing.
Can you see these two things in the animation below?
Recall that the general path description velocity and acceleration equations are given by the following:
v = v*et
a = v_dot*et + (v2/ρ)*en
Note that v_dot = dv/dt. For this problem, we do NOT know the speed as a function of time; instead, we know speed as a function of position, s. To find v_dot, we need to use the chain rule of differentiation: dv/dt = (dv/ds)*(ds/dt) = v*(dv/ds).