HOw to find acceleration when its equal to zero?

Question

The velocity v with arrow of a particle moving in the xy plane is given by v with arrow = (6.0t − 4.0t^2)i + 7.1j, with v with arrow in meters per second and t (> 0) is in seconds.

(a) What is the acceleration when t = 1.1 s? (Express your answer in vector form.)

I have -2.8i

(b) When (if ever) is the acceleration zero? (Enter 'never' if appropriate.)

(c) When (if ever) is the velocity zero? (Enter 'never' if appropriate.)

(d) When (if ever) does the speed equal 10 m/s? (Enter 'never' if appropriate.)

I don't understand the last three problems. Am I supposed to set the differentiated equation to zero/ten and then solve? But there's nothing left to solve...

Answer 1

B) when is the particle not accelerating in any direction. In other words when do we have,

$$v'(t)=0 \vec i +0 \vec j$$

Solution

$$(6-8t)\vec i +0 \vec j=0$$

$$6-8t=0$$

$$t=0.75 \text{seconds}$$

C) when is the car not in motion in any direction, in other words when is

$$v(t)=0 \vec i +0 \vec j$$

Solution:

The velocity is never $0$ because it he particle is always moving at a velocity of $7.1$ m/s in the $y$ direction. So there is no way for the velocity to be $0 \vec i+0 \vec j$.

D) when is the magnitude of the velocity vector equal to 10 m/s

In other words when is,

$$\sqrt{(6.0t-4.0t^2)^2+7.1^2}=10$$

Solution:

$$(6t-4t^2)^2+7.1^2=10^2$$

$$(6t-4t^2)^2=49.59$$

$$6t-4t^2= \pm \sqrt{49.59}$$

$$4t^2-6t \pm \sqrt{49.59}=0$$

Let's deal with

$$4t^2-6t+\sqrt{49.59}=0$$

First

By the quadratic formula,

$$t=\frac{6 \pm \sqrt{36-4(4)(\sqrt{49.59})}}{8}$$

Obviously this is no good so we only need to consider,

$$4t^2-6t-\sqrt{49.59}=0$$

$$t=\frac{6 +\sqrt{36+4(4)(\sqrt{49.59})}}{8}$$

Because we want $t \geq 0$.

This means $t \approx 2.27414$.