Question

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 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

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

Step-by-step 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 = $\frac{1}{0.05}$

t = 0 , t = 20

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

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

$\frac{dv}{dt} = 0$

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