Question

a particle moves along the x-axis so that at any time t>0, its velocity is given by v(t)=4-6t^2. if the particle is at a position x=7 at t=1 time, what is the position of the particle at time t=2?

Answer 1

-11.

Explanation:

We know that the velocity function v(t) is the derivative of the position function x(t).

So, we can integrate v(t) to find x(t) up to a constant of integration:

∫v(t) dt = ∫(4 - 6t^2) dt = 4t - 2t^3 + C

where C is the constant of integration.

We can find the value of C by using the initial condition that the particle is at position x=7 at t=1:

x(1) = 4(1) - 2(1)^3 + C = 7

C = 5

So, the position function is:

x(t) = 4t - 2t^3 + 5

To find the position of the particle at time t=2, we can substitute t=2 into the position function:

x(2) = 4(2) - 2(2)^3 + 5 = -11

Therefore, the position of the particle at time t=2 is -11.