15E - Applications to kinematics
The section builds upon the applications of differentiation to kinematics, where the mathematical relationship between position, velocity and acceleration was examined. For motion in a straight-line, we were able to calculate:
- Velocity from the position function: \(v(t)=x'(t)\).
- Acceleration from the velocity function: \(a(t)=v'(t)=x''(t)\).
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■ Remember: additional information is required to find the unique rules describing velocity and/or position. |
Distance and displacement are also related to the velocity function:
- The displacement of a particle is given by the signed area under the curve.
- The distance travelled by the particle is given by the absolute area under the curve.
15E - Example 1: Applications of integration to kinematics
A particle is travelling in a straight line. After 2 seconds, the velocity is \(-2\) \(m\) \(s^{-1}\) and the particle is \(3\) metres in the negative direction from some origin, \(O\). The acceleration, \(a\) \(m\) \(s^{-2}\), of the particle, after \(t\) seconds, is given by the equation \[a(t)=2\]
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15E - Example 1: Video Example
15E - Example 1: Practice
Question 1:
A particle is travelling in a straight line. The particle starts from rest at the origin, \(O\). The rule for the acceleration, \(a\) \(m\) \(s^{-2}\), of the particle is \[a(t)=-12t^2+24t\]
15E - Example 1: Solutions
Question 1: (a) \(v(t)=\int a(t)dt=-4t^3+12t^2\) (b) \(x(t)=\int v(t)dt=-t^4+4t^3\) (c) \(v(t)=0 \therefore t=3\) (d) \(x(3)=27\) metres |
15E - Example 2: Applications of integration to kinematics
A particle is travelling in a straight line. After 4 seconds, the particle has a velocity of \(-9\) \(m\) \(s^{-1}\) and the particle is \(3\) metres in the negative direction from some origin, \(O\). The acceleration, \(a\) \(m\) \(s^{-2}\), after \(t\) seconds is given by the equation \[a(t)=-6t+12\]
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15E - Example 2: Video Example
15E - Example 2: Practice
Question 1:
A particle is travelling in a straight line. After 1 second the particle is located 2 metres in the positive direction, relative to an origin \(O\). The velocity, \(v\) \(m\) \(s^{-1}\), is given by the rule \[v(t)=3t^2-6t\]
15E - Example 2: Solutions
Question 1: (a) \(v'(t)=0\therefore t=1 \therefore v(1)=-3\) \(m\) \(s^{-1}\) (b) \(x(t) = t^3-3t^2+4\) (c) \(x(0)=4\) (d) \(v(t)=24 \therefore t=4 \therefore x(4)=20\) (e) \(x(3)=4\) (f) Distance \(=8\) |
15E - Example 3: Applications of integration to kinematics (CAS)
A car is travelling along a stright road with a maximum speed limit of 110 km/h; however, the car only reaches a maximum velocity of 108 km/h. The velocity of the car, \(v\) \(m\) \(s^{-1}\), after \(t\) seconds, is given by the equation \[v(t)=10\sqrt{t+1}-18\]
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15E - Example 3: Video Example
15E - Example 3: Practice
Question 1 (TA): Need to fix!
The velocity of a speeding car, travelling in a straight line, is given by the equation \[v(t)=\frac{1500t}{(t+4)^2}\] where \(v\) is the velocity in kilometers per hour.
15E - Example 3: Solutions
Question 1: ABC |