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john is rollerblading down a long, straight path. at time zero, there is a mailbox about 1 m in front of him. in the 5 s time period that follows, john's velocity is given by the velocity versus time graph in the figure. taking the mailbox to mark the zero location, with positions beyond the mailbox as positive, plot his position versus time in the given position versus time graph.

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RevyBreeze

At time t=0 the mailbox is 1 m in front of John

therefore his initial position is  x_o = -1m

At time t=1s

x_1 = v_1 t_{1-0} + x_o

x_1 =(-2)(1) + (-1)

which gives us

3m

similarly at t=2s , x2 = -3m

at t=3s, x3 = -1m

at t=4s, x4 = 1m

at t=5s, x5 = 3m

the John's position at time t=4.61 seconds is given by

r = u (4.61-4 ) + 2.4

by plugging all the values we get

= 2(4.61-4) 1  

which gives us

2.22m

 

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