Example 9.4 -
Suppose while sitting in a parked car, you notice a
jogger approaching towards you in the side view mirror of R = 2 m. If
the jogger is running at a speed of 5 m s–1, how fast the image of the
jogger appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m,
and (d) 9 m away.
What I don't understand is the warped method they have solved this by...can't we simply differentiate the mirror formula wrt time and put in the values??
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1 Answers
1ya we can simply differentiate with time and get the answer
if we differentiate, we get dv/dt = -v2duu2dt
if u substitute the values u will find that the answers are almost the same
the one that u get through differentiation is more correct(it gives instantaneous velocity)
the answers in the textbook are average velocities(which is clearly mentioned there)
they just found out the change in u over 1 sec and hence the change in v,
then v(avg over 1 sec ) = change in v/ 1 sec
