Velocity of a particle varies with time as v=t^2-t-2. average speed of the particle in time interval from t=0 to t=4sec?

2 Answers
Jan 3, 2018

The average speed is #=4/3ms^-1#

Explanation:

The velocity as a function of time is

#v(t)=t^2-t-2#

The average speed #barv# in the time interval #t in [0,4]# is

#(4-0)barv=int_0^4(t^2-t-2)dt#

#4barv=[t^3/3-t^2/2-2t]_0^4#

#4barv=(4^3/3-4^2/2-2*4)-(0-0-0)#

#4barv=64/3-8-8=16/3#

#barv=4/3ms^-1#

Jan 3, 2018

3 m/s

No units for distance were given so I assumed meters.

Explanation:

The vt graph looks like this:

graph{x^2-x-2 [-10, 10, -5, 10]}

You can see that the particle is moving away from the origin at
t = 0 s. It then returns and moves away in a positive direction at
t = 2 s.

The question asks for the average speed which is a scalar quantity. This means we must find the total distance travelled as opposed to the displacement, which is a vector quantity.

To do this I will find the absolute value of the distance travelled between 0 and 2 s then add this to the absolute value of the distance travelled between 2 and 4 s.

In a vt graph the distance travelled is the area between the t axis and the line. We can find this by integration:

For the interval 0 to 2:

#sf(s_1=int_0^2v.dt)#

#sf(v=t^2-t-2)#

#:.##sf(s_1=int_0^2(t^2-t-2).dt)#

#sf(s_1=[t^3/3-t^2/2-2t]_0^2)#

#sf(s_1=(8/3-4/2-4)-(0))#

#sf(s_1=8/3-6=-10/3color(white)(x)m)#

For the interval 2 to 4:

#sf(s_2=int_2^4(t^2-t-2).dt)#

#sf(s_2=[t^3/3-t^2/2-2t]_2^4)#

#sf(s_2=(64/3-16/2-8)-(-10/3))#

#sf(s_2=16/3-(-10/3)=26/3color(white)(x)m)#

#sf(SigmaS=abs(s_1)+abs(s_2)=26/3+10/3=36/3=12color(white)(x)m)#

This is the total distance travelled.

#:.#average speed = total distance travelled / time taken =12/4 = 3 m/s