Anonymous
Anonymous asked in Science & MathematicsPhysics · 3 months ago

# Physics help?

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

Relevance
• 3 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)

• Ash
Lv 7
3 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