A 3kg mass experiences a force as shown in the graph below. It was initially at rest at the origin.
What is the velocity and position at t = 6s
- Andrew SmithLv 73 months ago
Impulse = change in momentum. F dt = m dv ie dv = F dt/ m (area of graph / m) so we can divide the motion into three regions. It gains velocity for three seconds, slows for the next two then remains at constant velocity.
In region 1 it gains 4*3/3 = 4 m/s. In the next region it loses -2*2/3= -4/3 m/s
giving a final velocity of 8/3 m/s
The distance moved in the first interval is average velocity * time = 4/2 *3 = 6m In the next interval the average velocity = (4+8/3)/2 for 2 seconds -> 20/3 m and for the final second it is 8/3 * 1 = 8/3 m
The total displacement is 18/3 + 20/3 +8/3 = 36/3 = 12 m
( and the average velocity is displacement / time = 12/2 = 2 m/s)
- AshLv 73 months ago
Average velocity = Area under the graph / mass of object
= [(4 x 3) + (-2 x 2)]/3 = (12 - 4)/3 = 8/3 m/s ≈ 2.67 m/s
position (at t=6) = Average velocity x time = (8/3)*6 = 16 m