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A vehicle is moving in a straight line. The velocity Vms^-1 at time t seconds after the vehicle starts is given by V = A (t− 0.05t^2) for 0 ≤ t ≤ 15. What is the value of A?

User WillEngler
by
8.1k points

2 Answers

6 votes

Answer:

Explanation:

User Fabrizio Mazzoni
by
8.8k points
5 votes

Answer:

The value of A is
(-\infty, 0) \cup(0, \infty)

Explanation:

Given:


V = A (t- 0.05t^2)

When velocity V= 0

we have


A (t- 0.05t^2) = 0


(t- 0.05t^2) = 0


t = 0.05t^2

t =0 , t =
(1)/(0.05)

t = 0 , t = 20

So when we draw a graph,The resulting graph will be a quadratic graph.


(dv)/(dt) = A[1 - 0.1t]


(dv)/(dt) = 0

t=10 sec

At t=10, the velocity will be


v(10) = [10 -0.05(10)^2]


v(10) = [10 -5]


v(10) = 5A

But according to the question A is not dependent on t

So A can be any value other than zero

User Julien Dumortier
by
9.0k points
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