Acceleration in 2D

Working with acceleration in two dimensions makes use of the vector addition and subtraction methods from the previous lessons. Recall the formula for acceleration:

    \[\vec{a} = \frac{\vec{v}_2 - \vec{v}_1}{\Delta t}\]

 

This formula can be used in two-dimensional problems to find acceleration if you are given \vec{v}_2, \vec{v}_1, and \Delta t.

 

Examples

Example 1

An airplane travelling at 240 m/s [28^\circ S of E] takes 35 s to change its velocity to 220 m/s [28^\circ E of S]. What is the average acceleration of the airplane over this time period?

 

Strategy:

  1. Use vector subtraction to find \vec{v}_2 - \vec{v}_1.
  2. Divide the result from Step 1 by \Delta t to find \vec{a}.

 

Solution

Step 1 for this problem type is to find \Delta\vec{v}. This is the same type of problem we solved in the last lesson. If this step confuses you go back and review the 2D Vector Subtraction lesson.

    \[[S][E]+\]

 

    \[\vec{v}_1 = 240 m/s [28^\circ S of E]\]

    \[\vec{v}_2 = 220 m/s [28^\circ E of S]\]

    \[\Delta t = 35 s\]

    \[\vec{a} = ?\]

2d-acceleration-1-1

2d-acceleration-1-2

 

    \[x\text{-components}\]

    \[y\text{-components}\]

    \[\vec{v}_{1x} = 240\cos 28\]

    \[\vec{v}_{1y} = 240\sin 28\]

    \[\vec{v}_{2x} = 220\sin 28\]

    \[\vec{v}_{2y} = 220\cos 28\]

    \[\begin{align*} \Delta\vec{v}_x &= \vec{v}_{2x} - \vec{v}_{1x} \\ &= 220\sin 28 - 240\cos 28 \\ &\approx -108.6 m/s\end{align*}\]

    \[\begin{align*} \Delta\vec{v}_y &= \vec{v}_{2y} - \vec{v}_{1y} \\ &= 220\cos 28 - 240\sin 28 \\ &\approx 81.6 m/s\end{align*}\]

 

2d-acceleration-1-3

 

    \[\begin{align*} \Delta v &= \sqrt{(\Delta v_x)^2 + (\Delta v_y)^2} \\ &= \sqrt{108.6^2 + 81.6^2} \\ &\approx 135.84 m/s\end{align*}\]

 

    \[\begin{align*} \theta &= \tan^{-1}\left(\frac{\Delta v_y}{\Delta v_x}\right) \\ &= \tan^{-1}\left(\frac{81.6}{108.6}\right) \\ &\approx 37^\circ\end{align*}\]

 

    \[\vec{v}_2 - \vec{v}_1 = 135.84 m/s [W37^\circ S]\]

 

For the second step, we will divide the above result by \Delta t. We will be dividing a vector by a scalar. This operation (and multiplying a vector and a scalar) can be done in the same way as dividing (or multiplying) two scalars. This operation will result in a vector, with the same direction as the vector you started with.

    \[\begin{align*} \vec{a} &= \frac{\vec{v}_2 - \vec{v}_1}{\Delta t} \\ &= \frac {135.84}{35} \\ &\approx 4 m/s^2 [W37^\circ S]\end{align*}\]

 

Example 2

A race car driver experiences an average acceleration of 0.15 m/s^2 [N25^\circ E] starting at a velocity of 45 m/s [S]. What is the velocity of the race car driver after 35 s?

 

In this question, we are given acceleration and asked to find the final velocity. We can rearrange our formula to solve for this.

    \[\vec{a} = \frac{\vec{v}_2 - \vec{v}_1}{\Delta t} = \vec{v}_2 = \vec{v}_1 + \vec{a}\Delta t\]

 

Strategy:

  1. Multiply \vec{a} by \Delta t.
  2. Use vector addition to add \vec{v}_1 and \vec{a\Delta t}.

 

Solution

    \[[N][E]+\]

 

    \[\vec{a} = 0.15 m/s^2 [N25^\circ E]\]

    \[\vec{v}_1 = 45 m/s [S]\]

    \[\Delta t = 35 s\]

    \[\vec{v}_2 = ?\]

 

    \[\begin{align*} \vec{a\Delta t} &= \vec{a}\Delta t \\ &= 0.15(35) \\ &= 5.25 m/s [N25^\circ E]\end{align*}\]

2d-acceleration-2-1

 

    \[x\text{-components}\]

    \[y\text{-components}\]

    \[\vec{v}_{1x} = 0\]

    \[\vec{v}_{1y} = -45 m/s\]

    \[\vec{a\Delta t}_x = 5.25\sin 25\]

    \[\vec{a\Delta t}_y = 5.25\cos 25\]

    \[\begin{align*} \vec{v}_{2x} &= \vec{v}_{1x} + \vec{a\Delta t}_x \\ &= 0 + 5.25\sin 25 \\ &\approx 2.22 m/s\end{align*}\]

    \[\begin{align*} \vec{v}_{2y} &= \vec{v}_{1y} + \vec{a\Delta t}_y \\ &= -45 + 5.25\cos 25 \\ &\approx -40.24 m/s\end{align*}\]

 

2d-acceleration-2-2

 

    \[\begin{align*} v_2 &= \sqrt{(v_{2x})^2 + (v_{2y})^2} \\ &= \sqrt{2.22^2 + 40.24^2} \\ &\approx 40 m/s\end{align*}\]

 

    \[\begin{align*} \theta &= \tan^{-1}\left(\frac{v_{2y}}{v_{2x}}\right) \\ &= \tan^{-1}\left(\frac{40.24}{2.22}\right) \\ &\approx 87^\circ\end{align*}\]

 

    \[\vec{v}_2 = 40 m/s [E87^\circ S]\]

 

Example 3

Jon walks through Central Park at 3.5 m/s [45^\circ E of N] when he realizes he is being followed. He changes his speed in 1.2 seconds to 4.5 m/s [12^\circ S of E]. Find his acceleration.

Video Solution

Solution

    \[[S][E]+\]

 

    \[\vec{v}_1 = 3.5 m/s [45^\circ E of N]\]

    \[\vec{v}_2 = 4.5 m/s [12^\circ S of E]\]

    \[\Delta t = 1.2 s\]

    \[\vec{a} = ?\]

2d-acceleration-3-1

2d-acceleration-3-2

 

    \[x\text{-components}\]

    \[y\text{-components}\]

    \[\vec{v}_{1x} = 3.5\sin 45\]

    \[\vec{v}_{1y} = - 3.5\cos 45\]

    \[\vec{v}_{2x} = 4.5\cos 12\]

    \[\vec{v}_{2y} = 4.5\sin 12\]

    \[\begin{align*} \Delta\vec{v}_x &= \vec{v}_{2x} - \vec{v}_{1x} \\ &= 4.5\cos 12 - 3.5\sin 45 \\ &\approx 1.93 m/s\end{align*}\]

    \[\begin{align*} \Delta\vec{v}_y &= \vec{v}_{2y} - \vec{v}_{1y} \\ &= 4.5\sin 12 - (-3.5\cos 45) \\ &\approx 3.41 m/s\end{align*}\]

 

2d-acceleration-3-3

 

    \[\begin{align*} \Delta v &= \sqrt{(\Delta v_x)^2 + (\Delta v_y)^2} \\ &= \sqrt{1.93^2 + 3.41^2} \\ &\approx 3.92 m/s\end{align*}\]

 

    \[\begin{align*} \theta &= \tan^{-1}\left(\frac{\Delta v_y}{\Delta v_x}\right) \\ &= \tan^{-1}\left(\frac{3.41}{1.93}\right) \\ &\approx 60^\circ\end{align*}\]

 

    \[\begin{align*} \vec{a} &= \frac{\vec{v}_2 - \vec{v}_1}{\Delta t} \\ &= \frac{3.92}{1.2} \\ &\approx 3.3 m/s^2 [60^\circ S of E]\end{align*}\]

 

Example 4

A car is travelling at 75 km/h [N26^\circ W], when it reaches a curve in the highway. if the car can accelerate at 2.1 m/s^2 [S15^\circ E] for 6.0 seconds, what is his final velocity?

Video Solution

Solution

    \[[N][W]+\]

 

    \[\begin{align*} \vec{v}_1 &= 75 km/h \times \frac{1000 m}{1 km} \times \frac{1 h}{3600 s} \\ &\approx 20.8 m/s [N26^\circ W]\end{align*}\]

    \[\vec{a} = 2.1 m/s^2 [S15^\circ E]\]

    \[\Delta t = 6.0 s\]

    \[\vec{v}_2 = ?\]

2d-acceleration-4-1

    \[\begin{align*} \vec{a\Delta t} &= 2.1(6.0) \\ &= 12.6 m/s [S15^\circ E]\end{align*}\]

2d-acceleration-4-2

 

    \[x\text{-components}\]

    \[y\text{-components}\]

    \[\vec{v}_{1x} = 20.8\sin 26\]

    \[\vec{v}_{1y} = 20.8\cos 26\]

    \[\vec{a\Delta t}_x = - 12.6\sin 15\]

    \[\vec{a\Delta t}_y = - 12.6\cos 15\]

    \[\begin{align*} \vec{v}_{2x} &= \vec{v}_{1x} + \vec{a\Delta t}_x \\ &= 20.8\sin 26 + (-12.6\sin 15) \\ &\approx 17.54 m/s\end{align*}\]

    \[\begin{align*} \vec{v}_{2y} &= \vec{v}_{1y} + \vec{a\Delta t}_y \\ &= 20.8\cos 26 + (-12.6\cos 15) \\ &\approx 6.52 m/s\end{align*}\]

2d-acceleration-4-3

    \[\begin{align*} v_2 &= \sqrt{(v_{2x})^2 + (v_{2y})^2} \\ &= \sqrt{17.54^2 + 6.52^2} \\ &\approx 18.71 m/s\end{align*}\]

 

    \[\begin{align*} \theta &= \tan^{-1}\left(\frac{v_{2y}}{v_{2x}}\right) \\ &= \tan^{-1}\left(\frac{6.52}{17.54}\right) \\ &\approx 20^\circ\end{align*}\]

\nbsp;

    \[\vec{v}_2 = 19 m/s [W20^\circ N]\]

 

Example 5

Dr. McCoy is walking from his quarters to the bridge at 2.2 m/s [N40^\circ E] (the Enterprise has a designated North). He is called to an emergency in the sick bay and accelerates at 1.15 m/s^2 [34^\circ N of W] to get there, what is his velocity when he arrives 12 seconds later?

Solution

    \[[N][E]+\]

 

    \[\vec{v}_1 = 2.2 m/s [N40^\circ E]\]

    \[\vec{a} = 1.15 m/s^2 [34^\circ N of W]\]

    \[\Delta t = 12 s\]

    \[\vec{v}_2 = ?\]

2d-acceleration-5-1

    \[\begin{align*} \vec{a\Delta t} &= 1.15(12) \\ &= 13.8 m/s [34^\circ N of W]\end{align*}\]

2d-acceleration-5-2

 

    \[x\text{-components}\]

    \[y\text{-components}\]

    \[\vec{v}_{1x} = 2.2\sin 40\]

    \[\vec{V}_{1y} = 2.2\cos 40\]

    \[\vec{a\Delta t}_x = - 13.8\cos 34\]

    \[\vec{a\Delta t}_y = 13.8\sin 34\]

    \[\begin{align*} \vec{v}_{2x} &= \vec{v}_{1x} + \vec{a\Delta t}_x \\ &= 2.2\sin 40 + (- 13.8\cos 34) \\ &\approx -10.0 m/s\end{align*}\]

    \[\begin{align*} \vec{v}_{2y} &= \vec{v}_{1y} + \vec{a\Delta t}_y \\ &= 2.2\cos 40 + 13.8\sin 34 \\ &\approx 9.4 m/s\end{align*}\]

 

2d-acceleration-5-3

 

    \[\begin{align*} v_2 &= \sqrt{(v_{2x})^2 + (v_{2y})^2} \\ &= \sqrt{10.0^2 + 9.4^2} \\ &\approx 13.7 m/s\end{align*}\]

 

    \[\begin{align*} \theta &= \tan^{-1}\left(\frac{v_{2y}}{v_{2x}}\right) \\ &= \tan^{-1}\left(\frac{9.4}{10.0}\right) \\ &\approx 43^\circ\end{align*}\]

 

    \[\vec{v}_2 = 14 m/s [W43^\circ N]\]

Additional Resources

  • Nelson Physics 12
    • Pg. 28 – 29

Assignment Questions

Assignment 1:

  • Pg. 29, #26
  • Pg. 31, #14