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.

I will consider each of the answer options below.

#=>#A: The magnitude of the x-coordinate of the particle is #4 "m"#.
#color(blue)(x(t)=(4t-t^2-4))#
#=>x(0)=(4(0)-(0)^2-4)#
#=>x=-4#
#=>#B: The magnitude of the average velocity is equal to the average speed for the time interval #t=0# to #t=2# seconds.
#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"#
#=>#C: The average acceleration is equal to the instantaneous acceleration during the time interval #t=0# to #t=2#.
#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.
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Answer 2

To find the velocity of the particle, differentiate the position function with respect to time:

v(t) = dx/dt = d(4t - t^2 - 4)/dt = 4 - 2t m/s

To find the acceleration of the particle, differentiate the velocity function with respect to time:

a(t) = dv/dt = d(4 - 2t)/dt = -2 m/s^2

To find the time(s) when the particle is at rest (velocity = 0), set v(t) = 0:

0 = 4 - 2t

Solve for t:

t = 2 s

To find the displacement of the particle at t = 2 s, substitute t = 2 into the position function:

x(2) = 4(2) - (2^2) - 4 = 4 m

Therefore, at t = 2 s, the particle is at rest and its displacement is 4 meters.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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