← Impulse and momentum change

Impulse from a force-time graph

The graph shows a monorail's velocity, steady from t = 2 s to t = 6 s. Calculate the monorail's displacement between t = 2 s and t = 6 s. Enter the value in meters, sign included.

time t (s)velocity v (m/s)0123456780+1+2+3+4+5+6+7+8+9

Haruki drags a loaded cart with a rope. The force from the rope is not steady: it starts at 0 N and climbs to 40 N over 0.3 s.

The net force on the cart keeps changing. How much impulse does a changing force deliver to the cart?

time t (s)net force F (N)0.10.20.30.41020304050triangle
The climbing force on Haruki's cart. The shaded triangle's area is the impulse delivered.

Here is the net force on Haruki's cart, drawn on a force-time graph.

The graph shows the net force at each moment: 0 N at the start, climbing steadily to 40 N at 0.3 s.

J = FΔt works only while F holds one value, so a changing force needs a new reading.

Cut the time into thin slices, so the force barely changes inside each one.

Each slice's impulse is its force times its thin width of time. On the graph, that product is the area of a thin strip.

Add every strip, and the total is the whole area under the line.

So the impulse delivered equals the area under the force-time graph.

You have read an area like this before: the area under a velocity-time graph gave the displacement.

The reading move is the same; only the quantities changed.

For straight-line graphs, the area is a shape you know: a rectangle, or a triangle.

The impulse delivered equals the area under the force-time graph.

The net force on Haruki's cart climbs steadily from 0 N to 40 N over 0.3 s. Determine the impulse delivered.

time t (s)net force F (N)0.10.20.30.41020304050triangle
The climbing force again: the shaded region is a triangle.

Write down what the graph shows:

The force climbs from 0 N to 40 N over 0.3 s, so the shaded region is a triangle.

Write down the area formula:

area = ½ × base × height

Substitute in the values, and calculate:

height × base: 40 × 0.3 = 12, so J = 0.5 × 12 = 6

The impulse delivered is 6 N·s.

The same cart on a second trip: the rope now holds a steady 25 N for 0.4 s. Determine the impulse delivered.

time t (s)net force F (N)0.10.20.30.40.510202530rectangle
The second trip's steady force. The shaded rectangle's area is the impulse delivered.

Write down what the graph shows:

The force holds at 25 N from 0 to 0.4 s, so the shaded region is a rectangle.

Write down the area formula:

area = height × base

Substitute in the values, and calculate:

J = 25 × 0.4 = 10

The impulse delivered is 10 N·s.

The graph shows the net force on a wagon: a steady 30 N from 0 to 0.5 s. The shaded region is a rectangle. Enter the impulse delivered in N·s.

time t (s)net force F (N)0.10.20.30.40.50.610203040shaded region

The graph shows the net force on a skateboard: it climbs steadily from 0 N to 50 N over 0.2 s. The shaded region is a triangle. Enter the impulse delivered in N·s.

time t (s)net force F (N)0.050.10.150.20.250.3102030405060shaded region

The graph shows the net force on a dart as it leaves a launcher: the force starts at 90 N and falls steadily to 0 N over 0.08 s. The shaded region is a triangle. Enter the impulse delivered in N·s.

time t (s)net force F (N)0.020.040.060.080.1102030405060708090100shaded region

Your turn

The graph shows the net force on a canoe: a steady 45 N from 0 to 0.6 s. Enter the impulse delivered in N·s.

time t (s)net force F (N)0.10.20.30.40.50.60.710203040455060shaded region

The graph shows the net force on a drone as a gust builds: it climbs steadily from 0 N to 24 N over 1.5 s. Enter the impulse delivered in N·s.

time t (s)net force F (N)0.511.5210202430shaded region

The graph shows the net force on a scooter: a steady 20 N from 0 to 0.4 s, then falling steadily to 0 N at 0.7 s. Enter the total impulse delivered in N·s.

time t (s)net force F (N)0.10.20.30.40.50.60.70.8102030shaded region