Figure is missing: find it in attachment.
Answer:
+4.50 m/s
Explanation:
The mass of the cart is
m = 40.0 kg
and its initial velocity is
u = -3.00 m/s (negative x-direction)
The graph represents the force applied to the cart versus time. We know that the area under the graph corresponds to the impulse, which is the change in momentum of the cart. We also notice that the force applied in the two parts is constant, so we can write the impulse simply as product between the force applied and the time interval:
[tex]\Delta p = F \Delta t[/tex]
In the first part of the motion (from 0 s to 20 s), the force applied is F = +20 N, so the impulse (and the change in momentum) is
[tex]\Delta p_1 = (+20 N)(20 s - 0 s ) =+400 Ns[/tex]
In the second part of the motion (from 20 s to 30 s), the force applied is
F = -10 N
So the impulse is
[tex]\Delta p_2 = (-10 N)(30 s-20 s)=-100 Ns[/tex]
So the total change in momentum of the cart is
[tex]\Delta p = +400 - 100 = +300 Ns[/tex]
The change in momentum can be written as
[tex]\Delta p = m(v-u)[/tex]
where v is the final velocity of the cart. Solving for v,
[tex]v=u+\frac{\Delta p}{m}=-3.00 + \frac{+300}{40.0}=+4.50 m/s[/tex]
So, 4.50 m/s in the positive direction.
