Question

A motorcyclist heading east accelerates after he passes the signpost marking the city limits. His acceleration is constant at40 m/s^2. At time t=0, he is 5.0m E of the signpost moving east at 15 m/s.

a) Find his position and velocity at t=2.0
b) Where is the motorcycle when his velocity is 25m/s

please use the following formulas that are needed to solve this:
d=vi+1/2at^2
vf^2=vi^2+2ad
vf=vi+at
d=((vf+v1)/2)t
xf-xi=vi+1/2at^2
xf: final position, xi: initial position

Thank you!

Answer 1

his position is given by \[x _{f}=x _{i}+v _{i}t+1/2at^{2}\]your last equation. Plug in the numbers\[x _{f}=5.0m+(15m/s)(2s)+1/2(40m/s^{2})(2s)^{2}\]To find his position use your second formula\[v _{f}^{2}=v _{i}^{2}+2a \Delta x\]and rearrange to find x final\[x _{f} - x _{i} =\frac {v _{f}^{2}-v _{i}^{2}} {2a}\]\[x _{f} = x _{i} +\frac {v _{f}^{2}-v _{i}^{2}} {2a}\]and plug in\[x _{f}=5.0m+\frac {25m/s^2-15m/s^{2}} {2(40m/s^{2})}\]

Answer 2

Sorry - the formatting on the last equation is wrong...it should be\[x _{f} =5.0m + \frac {(25m/s)^{2} -(15m/s)^{2}} {2(40m/s^{2})}\]

Answer 3

His velocity at 2 s is given by your 3rd formula\[v _{f}=v _{i} +at=15m/s+(40m/s^{2})(2s)\]

Answer 4

thank you! <3