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.

1 Answer
Aug 18, 2017

Answer:

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.