Amass m is at rest on a horizontal frictionless surface at t=0. then a constant force f0 acts on it for a time t0. suddenly the force doubles to 2f0 and remains constant until t=2t0. part a determine the total distance traveled from t=0 to t=2t0.

The total distance travel by the object is 2.5(F0/m)(t0)^2

Since constant force is applied on the object so we can apply the equation of linear motion.

According to law of inertia

F=ma

therfore for the time period t=0 to t_o the acceleration of the object is

a=F0/m

And the velocity of the object after t0 sec is calculated by the first equation of motion

v=u+at

since the initial velocity of the object is 0, therefore the velocity of the object after t0 sec is

v=(F0/m)*t0

and the distance cover by the object during t0 sec is calculated by the second equation of motion

s1=ut+0.5at^2

Plugging the values

s1=0.5(F0/m)(t0)^2

Now the acceleration of the object during 2t0 sec is

a2=2F0/m

therefore the distance travel by the object from time t0 to 2t0 is calculated as

s2=(F0/m)*t0*t0+0.5(2F0/m)*(t0)^2

=(2F0/m)(t0)^2

Therefore total distance travel by the object is

S=s1+S2

=2.5(F0/m)(t0)^2

the area between the graph and the t axis