AI HL · AHL 4
0 of 160 skills done
Variable velocity in two dimensions
AHL 4.23
Screenshot and video to come
Screenshot and video to come
Screenshot and video to come
The velocity of an object is \(\mathbf{v} = \begin{pmatrix}3t^{2} - 4\\8 - 5t\end{pmatrix}\) m s⁻¹. At \(t = 0\) it is at \(\mathbf{r}_0 = \begin{pmatrix}2\\-1\end{pmatrix}\) m. Find (a) the acceleration when \(t = 2\), (b) the position vector when \(t = 4\).
Have a go first. Then open the working one line at a time.
\(\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = \begin{pmatrix}6t\\-5\end{pmatrix}\)
Differentiate each component separately.
\(\mathbf{a}(2) = \begin{pmatrix}12\\-5\end{pmatrix}\) m s⁻²
Put t = 2 into both components.
\(\mathbf{r} = \begin{pmatrix}t^{3} - 4t + c_1\\-\frac{5t^{2}}{2} + 8t + c_2\end{pmatrix}\)
Integrate each component and keep a constant for each.
\(c_1 = 2,\ c_2 = −1\)
At t = 0 the position is (2, −1). The t terms vanish, so each constant is the starting value.
\(\mathbf{r}(4) = \begin{pmatrix}50\\-9\end{pmatrix}\) m
Put t = 4 into the full expression for r.
The velocity of a particle is \(\mathbf{v} = \begin{pmatrix}10e^{2t}\\8e^{-4t}\end{pmatrix}\) m s⁻¹. At \(t = 0\) it is at \(\mathbf{r}_0 = \begin{pmatrix}3\\1\end{pmatrix}\) m. Find the acceleration and the position vector at time \(t\), then the position vector when \(t = 0.5\).
Have a go first. Then open the working one line at a time.
\(\mathbf{a} = \begin{pmatrix}20e^{2t}\\-32e^{-4t}\end{pmatrix}\)
The derivative of \(e^{kt}\) is \(ke^{kt}\): multiply by the 2 and by the −4.
\(\mathbf{r} = \begin{pmatrix}5e^{2t} + c_1\\-2e^{-4t} + c_2\end{pmatrix}\)
The integral of \(e^{kt}\) is \(e^{kt}\) divided by \(k\).
\(c_1 = 3 - 5 = −2,\ c_2 = 1 + 2 = 3\)
At t = 0, \(e^0 = 1\), so each exponential term gives its coefficient.
\(\mathbf{r}(0.5) = \begin{pmatrix}11.6\\2.73\end{pmatrix}\) m
Put t = 0.5 into r.
The velocity of a particle is \(\mathbf{v} = \begin{pmatrix}9\cos 3t\\-6\sin 3t\end{pmatrix}\) m s⁻¹. At \(t = 0\) it is at \(\mathbf{r}_0 = \begin{pmatrix}0\\5\end{pmatrix}\) m. Find the acceleration and the position vector at time \(t\), then the i-component of the acceleration when \(t = 1\).
Have a go first. Then open the working one line at a time.
\(\mathbf{a} = \begin{pmatrix}-27\sin 3t\\-18\cos 3t\end{pmatrix}\)
The derivative of cos 3t is −3 sin 3t, and of sin 3t is 3 cos 3t. The chain rule brings out the 3.
\(\mathbf{r} = \begin{pmatrix}3\sin 3t + c_1\\2\cos 3t + c_2\end{pmatrix}\)
Divide by the 3 when you integrate.
\(c_1 = 0 - 0 = 0,\ c_2 = 5 - 2 = 3\)
At t = 0, sin 0 = 0 and cos 0 = 1.
From the calculator, in radians: \(a_1(1) = −27\sin 3 = −3.81\) m s⁻²
Radian mode for time in seconds.
The velocity of an object at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix}2t^{2} - 9\\6 - t\end{pmatrix}\) m s⁻¹. Find its speed after 3 seconds.
Have a go first. Then open the working one line at a time.
\(\mathbf{v}(3) = \begin{pmatrix}9\\3\end{pmatrix}\)
Substitute first.
\(|\mathbf{v}| = \sqrt{9^2 + 3^2} = \sqrt{90} = 9.49\) m s⁻¹
Speed is the magnitude of the velocity vector. It is a number, not a vector.
The velocity of a particle is \(\mathbf{v} = \begin{pmatrix}4e^{3t}\\9e^{-t}\end{pmatrix}\) m s⁻¹. Find its speed when \(t = 0.3\).
Have a go first. Then open the working one line at a time.
From the calculator: \(\mathbf{v}(0.3) = \begin{pmatrix}9.84\\6.67\end{pmatrix}\)
Evaluate each component.
\(|\mathbf{v}| = \sqrt{9.84^2 + 6.67^2} = 11.9\) m s⁻¹
Both components count, even though the i-component is growing.
The velocity of a particle is \(\mathbf{v} = \begin{pmatrix}7\cos 2t\\5\sin 2t\end{pmatrix}\) m s⁻¹. Find its speed when \(t = 1.5\).
Have a go first. Then open the working one line at a time.
From the calculator, in radians: \(\mathbf{v}(1.5) = \begin{pmatrix}-6.93\\0.706\end{pmatrix}\)
2t = 3 radians.
\(|\mathbf{v}| = \sqrt{(−6.93)^2 + 0.706^2} = 6.97\) m s⁻¹
Square each component, add, take the square root.
The position vector of a skater at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix}4t^{2} - t\\2\cos 2t\end{pmatrix}\) m. Find the speed of the skater when \(t = 1.5\).
Have a go first. Then open the working one line at a time.
\(\mathbf{v} = \dfrac{d\mathbf{r}}{dt} = \begin{pmatrix}8t - 1\\-4\sin 2t\end{pmatrix}\)
Velocity is the derivative of position.
From the calculator, in radians: \(\mathbf{v}(1.5) = \begin{pmatrix}11\\-0.564\end{pmatrix}\)
Substitute t = 1.5 into the velocity, not the position.
\(|\mathbf{v}| = 11.0\) m s⁻¹
Speed is the magnitude of v.
The velocity of an object is \(\mathbf{v} = (5 + 2t)\mathbf{i} + (11 - 3t)\mathbf{j}\) m s⁻¹. (a) Find the angle the initial velocity makes with \(\mathbf{i}\). (b) Show that the acceleration is constant and find its magnitude.
Have a go first. Then open the working one line at a time.
\(\mathbf{v}(0) = 5\mathbf{i} + 11\mathbf{j}\)
Initial means t = 0.
\(\tan\theta = \dfrac{11}{5},\ \theta = 65.6^\circ\)
Sketch the right-angled triangle: 5 along i and 11 along j.
\(\mathbf{a} = 2\mathbf{i} - 3\mathbf{j}\)
The acceleration has no t in it, so it is constant.
\(|\mathbf{a}| = \sqrt{2^2 + (−3)^2} = \sqrt{13} = 3.61\) m s⁻²
Constant in size and direction.
The displacement of a particle from the origin is \(\mathbf{r} = \begin{pmatrix}t^{3} - 2t\\12 - 3t^{2}\end{pmatrix}\) m at time \(t\) seconds. Find its distance from the origin when \(t = 3\).
Have a go first. Then open the working one line at a time.
\(\mathbf{r}(3) = \begin{pmatrix}21\\-15\end{pmatrix}\)
Displacement at t = 3.
\(|\mathbf{r}| = \sqrt{21^2 + (−15)^2} = \sqrt{666} = 25.8\) m
Distance is the magnitude of the displacement from the origin. Compare with speed, the magnitude of v.