# A particle is moving in x - axis according to relation x= (4t - t^2 - 4) m then? **The question has multiple answers**.

## A: magnitude of x coordinate of particle is 4m B: magnitude of a average velocity is equal to average speed, for time interval t=0 to t=2sec C: average acceleration is equal to instantaneous acceleration during interval t=0 to t=2sec D: distance traveled in interval t=0 to t=4sec is 8m. The question has multiple answers.

Aug 18, 2017

I got A,B, C, and D are correct.

#### Explanation:

I will consider each of the answer options below.

$\implies$A: The magnitude of the x-coordinate of the particle is $4 \text{m}$.

• I assume this refers to the position of the particle at $t = 0$, since it travels along the x-axis only. We can substitute $0$ into the given equation for position and see what we get for x.

$\textcolor{b l u e}{x \left(t\right) = \left(4 t - {t}^{2} - 4\right)}$

$\implies x \left(0\right) = \left(4 \left(0\right) - {\left(0\right)}^{2} - 4\right)$

$\implies x = - 4$

• Therefore, the initial position of the particle is $- 4 \text{m}$, and $\left\mid - 4 \right\mid = 4$, so answer A is correct.

$\implies$B: The magnitude of the average velocity is equal to the average speed for the time interval $t = 0$ to $t = 2$ seconds.

• We can compute the average velocity and average speed independently and compare the values we obtain.
• The most important distinction between the two quantities is that the average speed is concerned with the distance the object travels over a given time period, whereas the average velocity is concerned with the displacement of the object over a given time period.

$\textcolor{b l u e}{{v}_{\text{avg}} = \frac{\Delta x}{\Delta t}}$

We will first calculate the displacement $\Delta x$, where $\Delta x = {x}_{f} - {x}_{i}$.

${x}_{i} = x \left(0\right) = - 4 \text{m}$

${x}_{f} = x \left(2\right) = 0 \text{m}$

Therefore, $\Delta x = 0 - \left(- 4\right) = 4 \text{m}$.

We are given $\Delta t = {t}_{f} - {t}_{i} = 2 - 0 = 2 \text{s}$

v_"avg"=(4"m")/(2"s")

${v}_{\text{avg"=2"m"//"s}}$ (in the direction of the positive x-axis)

Now we calculate the average speed:

$\textcolor{b l u e}{{s}_{\text{avg}} = \frac{\Delta d}{\Delta t}}$

• At $t = 0$, the particle is at $x \left(0\right) = - 4 \text{m}$

• At $t = 1$, the particle is at $x \left(1\right) = - 1 \text{m}$

• At $t = 2$, the particle is at $x \left(2\right) = 0 \text{m}$

So, the object traveled a total distance of $\left(3 + 1\right) \text{m"=4"m}$

Therefore, $\Delta d = 4 \text{m}$ and the time period is still $2 \text{s}$

s_"avg"=(4"m")/(2"s")

${s}_{\text{avg"=2"m"//"s}}$

• Therefore, ${v}_{\text{avg"=s_"avg}}$ and answer B is correct.

$\implies$C: The average acceleration is equal to the instantaneous acceleration during the time interval $t = 0$ to $t = 2$.

• We can find the instantaneous acceleration by taking the second derivative of the given equation for position. The average acceleration can be found as the change in velocity over time.

$\textcolor{b l u e}{{a}_{\text{avg}} = \frac{\Delta v}{\Delta t}}$

We will first need to calculate $\Delta v$, the change in velocity. We can find the velocity by taking the first derivative of the given equation for position and using the given time interval.

$v \left(t\right) = x ' \left(t\right) = 4 - 2 t$

${v}_{i} = v \left(0\right) = 4$

${v}_{f} = v \left(2\right) = 0$

Therefore, $\Delta v = 0 - 4 = - 4 \text{m"//"s}$.

a_"avg"=(-4"m"//"s")/(2"s")

${a}_{\text{avg"=-2"m"//"s}}^{2}$

We can find the instantaneous acceleration by taking the derivative of the equation for velocity that we derived above, which is the second derivative of position.

$a \left(t\right) = v ' \left(t\right) = - 2 {\text{m"//"s}}^{2}$

We see that $\vec{a} = {a}_{\text{avg}}$ and therefore answer C is correct.

$\implies$D: The distance traveled in the interval $t = 0$ to $t = 4$ seconds is $8 \text{m}$.

We found above that the distance traveled by the particle when $t \in \left[0 , 4\right]$ was $4 \text{m}$, so we know that at $x \left(2\right)$, the particle has already traveled $4 \text{m}$.

• At $t = 2$, the particle is at $x \left(2\right) = 0 \text{m}$

• At $t = 3$, the particle is at $x \left(3\right) = - 1 \text{m}$

• At $t = 4$, the particle is at $x \left(4\right) = 4 \text{m}$

So the particle travels an additional $\left(1 + 3\right) = 4 \text{m}$. Therefore, the total distance is $4 + 4 = 8 \text{m}$, and answer D is correct.