Position, velocity and acceleration in component form.
Example 1
A particle has position r(t)=(6t2,5t3). Find the y−component of acceleration at t=4.
Follow the position curve. The gold ring keeps the question time fixed. Axes use different scales; the tangent shows direction, not speed.Show worked answer
Answer: 120
Differentiate each component: v(t)=(12t,15t2).
Differentiate again: a(t)=(12,30t).
At t=4, the y−component is 120.
Example 2
A particle has velocity v(t)=(20t,7). Its initial position is (4,0). Find its x−coordinate at t=6.
Mint is horizontal velocity; lavender is vertical velocity. Initial x = 4. Optional mint shading represents horizontal displacement, not position. Axes use different scales.Show worked answer
Answer: 364
Integrate the x−component: x(t)=10t2+C.
The initial position gives C=4.
At t=6,x=364.
Example 3
A particle has position r(t)=(7t2,6t3). Find the y−component of acceleration at t=5.
Follow the position curve. The gold ring keeps the question time fixed. Axes use different scales; the tangent shows direction, not speed.Show worked answer
Answer: 180
Differentiate each component: v(t)=(14t,18t2).
Differentiate again: a(t)=(14,36t).
At t=5, the y−component is 180.
Vector Differentiation & Integration Practice | Year 12 Specialist | Ascent Arcade