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Part H)
At t = 0 a truck starts from rest at x = 0 and speeds up in the positive x -direction on a straight road with acceleration aT . At the same time, t = 0, a car is at x = 0 and traveling in the positive x -direction with speed vC . The car has a constant negative x -acceleration: acar−x=−aC , where aC is a positive quantity. At what time does the truck pass the car? Express your answer in terms of the variables vC , aC , and aT .

Part I)
At what coordinate does the truck pass the car?
Express your answer in terms of the variables vC , aC , and aT .



Answer :

okpalawalter8

At t = 0 a truck starts from rest at x = 0 and speeds up in the positive x-direction on..., the time the truck passes the car and the coordinate the truck passes the car is mathematically given as

  • [tex]t = \frac{2v_{C}}{a_{T} + a_{C}}[/tex]
  • [tex]d_{T} = v_{T}t + \frac{a_{T}t^{2}}{2}[/tex]

What time does the truck pass the car and coordinate does the truck pass the car?

Generally, the equation for Distance traveled is mathematically given as

By Truck

[tex]d_{T} = v_{T}t + \frac{a_{T}t^{2}}{2}[/tex]

By Car

[tex]d_{C} = v_{C}t - \frac{a_{C}t^{2}}{2}[/tex]

In conclusion, when the conditions the truck passes the car

[tex]d_{T} = d_{C}[/tex]

Therefore

[tex]v_{t} + \frac{a_{t}t}{2} = v_{C} - \frac{a_{C}t}{2}[/tex]

[tex]\frac{a_{T}t}{2} = v_{C} - \frac{a_{C}t}{2}[/tex]

Therefore

[tex]t = \frac{2v_{C}}{a_{T} + a_{C}}[/tex]

differentiation of  T

[tex]d_{T} = v_{T}t + \frac{a_{T}t^{2}}{2}[/tex]

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