Calculus question Help

Question

A particle moves along the y-axis so that at time t is greater than or equal to 0 its position is given by y(t)=t^3-4t^2+4t+3. Is the particle moving down the y-axis with increasing velocity at time t=1?

Answer 1

To determine whether the particle is moving down the y-axis with increasing velocity at time t=1, we can find the velocity of the particle at that time using the formula for velocity:

v(t) = d/dt(y(t)) = 3t^2 - 8t + 4

Plugging in t=1, we get:

v(1) = 31^2 - 81 + 4 = -1

Since the velocity at time t=1 is negative, the particle is moving down the y-axis. However, to determine whether the velocity is increasing at that time, we need to find the acceleration of the particle at t=1. We can do this using the formula for acceleration:

a(t) = d/dt(v(t)) = 6t - 8

Plugging in t=1, we get:

a(1) = 6*1 - 8 = -2

Since the acceleration at time t=1 is negative, the velocity of the particle is decreasing at that time. Therefore, the particle is not moving down the y-axis with increasing velocity at time t=1.

Answer 2

We have been told that the particle is moving along the y-axis. This is our spatial dimension.

The variable t is our time dimension.

The position is given by: y(t) = t^3-4t^2+4t+3

The change in position over the change in time is velocity, or the first derivative.

The change in velocity over the change in time, is acceleration, or the second derivative. This is what we are being asked to find at t = 1.

We employ the power rule to find:

y(t) = t^3-4t^2+4t+3

First derivative: y'(t) = 3t^2 - 8t + 4

Second derivative: y''(t) = 6t^2 - 8

Evaluating at t = 1 yields:

y''(1) = 6(1)^2 -8

y''(1) = 6 - 8

y''(1) = -2

So the answer is "The change in velocity over the change in time at t = 1 is -2, which is decreasing in the positive y, or increasing down in the negative Y direction. "