Asked by sw
A car travels along the curved path with radius of curvature p= 30m such that its speed is increased by at = (0.5e^t) m/s^2, where t is in seconds. determine the magnitudes of its velocity and acceleration after the car has traveled s = 18m starting from rest. neglect the size of the car.
Answers
Answered by
drwls
The distance s travelled in time t' is
S (t') = Double integral a_t dt
t=0 to t'
18 = 0.5*e^t' meters
t' = ln 36 = 3.584 s
V (t') = Single integral a_t dt
t=0 to t'
= 0.5*e^t' = 18 m/s
a_t (t') = 18 m/s^2
There is also a centrifugal acceleration component,
a_r = V(t')^2/R = 324/30 = 10.8 m/s^2
S (t') = Double integral a_t dt
t=0 to t'
18 = 0.5*e^t' meters
t' = ln 36 = 3.584 s
V (t') = Single integral a_t dt
t=0 to t'
= 0.5*e^t' = 18 m/s
a_t (t') = 18 m/s^2
There is also a centrifugal acceleration component,
a_r = V(t')^2/R = 324/30 = 10.8 m/s^2
Answered by
sw
you're a lifesaver
thank you!
thank you!
Answered by
Kyle
Above answer is completely wrong.. it seems he forgot that it's an initial value problem
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