Question

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.

Answer 1

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

Explanation:

I will consider each of the answer options below.

`=>`A: The magnitude of the x-coordinate of the particle is `4 "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.

`color(blue)(x(t)=(4t-t^2-4))`

`=>x(0)=(4(0)-(0)^2-4)`

`=>x=-4`

• Therefore, the initial position of the particle is `-4"m"`, and `abs(-4)=4`, so answer A is correct.

`=>`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.

`color(blue)(v_"avg"=(Deltax)/(Deltat))`

We will first calculate the displacement `Deltax`, where `Deltax=x_f-x_i`.

`x_i=x(0)=-4"m"`

`x_f=x(2)=0"m"`

Therefore, `Deltax=0-(-4)=4"m"`.

We are given `Deltat=t_f-t_i=2-0=2"s"`

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

`v_"avg"=2"m"//"s"` (in the direction of the positive x-axis)

Now we calculate the average speed:

`color(blue)(s_"avg"=(Deltad)/(Deltat))`

• At `t=0`, the particle is at `x(0)=-4"m"`

• At `t=1`, the particle is at `x(1)=-1"m"`

• At `t=2`, the particle is at `x(2)=0"m"`

So, the object traveled a total distance of `(3+1)"m"=4"m"`

Therefore, `Deltad=4"m"` and the time period is still `2"s"`

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

`s_"avg"=2"m"//"s"`

• Therefore, `v_"avg"=s_"avg"` and answer B is correct.

`=>`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.

`color(blue)(a_"avg"=(Deltav)/(Deltat))`

We will first need to calculate `Deltav`, 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(t)=x'(t)=4-2t`

`v_i=v(0)=4`

`v_f=v(2)=0`

Therefore, `Deltav=0-4=-4"m"//"s"`.

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

`a_"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(t)=v'(t)=-2"m"//"s"^2`

We see that `veca=a_"avg"` and therefore answer C is correct.

`=>`D: The distance traveled in the interval `t=0` to `t=4` seconds is `8"m"`.

We found above that the distance traveled by the particle when `tin[0,4]` was `4"m"`, so we know that at `x(2)`, the particle has already traveled `4"m"`.

• At `t=2`, the particle is at `x(2)=0"m"`

• At `t=3`, the particle is at `x(3)=-1"m"`

• At `t=4`, the particle is at `x(4)=4"m"`

So the particle travels an additional `(1+3)=4"m"`. Therefore, the total distance is `4+4=8"m"`, and answer D is correct.