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2 years ago

Part H)

1 answer:

5 0

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

$t = \frac{2v_{C}}{a_{T} + a_{C}}$
$d_{T} = v_{T}t + \frac{a_{T}t^{2}}{2}$

<h3>What time does the truck pass the car and coordinate does the truck pass the car?</h3>

Generally, the equation for Distance traveled is mathematically given as

By Truck

$d_{T} = v_{T}t + \frac{a_{T}t^{2}}{2}$

By Car

$d_{C} = v_{C}t - \frac{a_{C}t^{2}}{2}$

In conclusion, when the conditions the truck passes the car

$d_{T} = d_{C}$

Therefore

$v_{t} + \frac{a_{t}t}{2} = v_{C} - \frac{a_{C}t}{2}$

$\frac{a_{T}t}{2} = v_{C} - \frac{a_{C}t}{2}$

Therefore

$t = \frac{2v_{C}}{a_{T} + a_{C}}$

differentiation of T

$d_{T} = v_{T}t + \frac{a_{T}t^{2}}{2}$

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Answer:

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Explanation:

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