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?

Answer :

nandhini123

Answer:

The value of A is [tex](-\infty, 0) \cup(0, \infty)[/tex]

Step-by-step explanation:

Given:

[tex]V = A (t- 0.05t^2)[/tex]

When velocity V= 0

we have

[tex]A (t- 0.05t^2) = 0[/tex]

[tex](t- 0.05t^2) = 0[/tex]

[tex]t = 0.05t^2[/tex]

t =0 , t = [tex]\frac{1}{0.05}[/tex]

t = 0 , t = 20

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

[tex]\frac{dv}{dt} = A[1 - 0.1t][/tex]

[tex]\frac{dv}{dt} = 0[/tex]

t=10 sec

At t=10, the velocity will be

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

[tex]v(10) = [10 -5][/tex]

[tex]v(10) = 5A[/tex]

But according to the question A is not dependent on t

So A can be any  value other than zero

Answer:

Step-by-step explanation:

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