Question #175019

A car stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by xt=bt2-ct3 where, b=2.40m/s2 and c=0.120m/s3


A.) Calculate the average velocity of the car for the time interval t=0s to t=10.0s


B.) Calculate the instantaneous velocity of the car at t=0s; t=5.0s; and t=10.0s



1
Expert's answer
2021-03-30T06:44:06-0400

Answer

A. distance from the light is given by x(t)=bt2ct3x(t) =bt^2-ct^3

where, b=2.40m/s22.40m/s^2 and c=0.120m/s30.120m/s^3

distance from the light is given by

x(t)=bt2ct3x(t) =bt^2-ct^3 at t=0

x=0 m

At t=10sec

x(t)=2.40×1000.12×1000x(t) =2.40\times 100-0.12\times1000

=120m

the average velocity of the car for the time interval t=0s to t=10.0s

Va=dxdt=1200100=12m/sV_a=\frac{dx}{dt}\\=\frac{120-0}{10-0}\\=12m/s

B. the instantaneous velocity of the car at t=0s; t=5.0s; and t=10.0s

v=dxdt=2bt3ct2v=\frac{dx}{dt}=2bt-3ct^2

So at t=0 v=0

At t=5sec

v=2×2.4×53×0.12×25=15m/sv=2\times2.4\times5-3\times0.12\times25\\=15m/s

At t=10

v=2×2.4×103×0.12×100=12m/sv=2\times2.4\times10-3\times0.12\times100\\=12m/s





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Comments

MUHAMMED BABANGIDA TALASE
22.06.21, 14:07

the guy who answer this was expert on physics thank you very much

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