A person walks 20 meters north, then turns 30 degrees east of norh and walks another 15 meters

A person walks 20 meters north, then turns 30 degrees east of norh and walks another 15 meters. Calculate the resultant displacement using trigonometric identities?

The correct answer and explanation is:

To calculate the resultant displacement, we can break the motion into its x (east-west) and y (north-south) components and then combine them.

Step 1: Break the movement into components

1. The person first walks 20 meters north. This motion is purely along the y-axis, so the x-component is 0 and the y-component is 20 meters.
   • x1=0x_1 = 0
   • y1=20y_1 = 20
2. The person then turns 30° east of north and walks 15 meters. This displacement has both x and y components.
   • To find the x and y components, we use trigonometric functions:
     • The x-component is 15⋅sin⁡(30∘)15 \cdot \sin(30^\circ).
     • The y-component is 15⋅cos⁡(30∘)15 \cdot \cos(30^\circ).
   Using known values:
   • sin⁡(30∘)=0.5\sin(30^\circ) = 0.5
   • cos⁡(30∘)=32≈0.866\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866
   So:
   • x2=15⋅0.5=7.5x_2 = 15 \cdot 0.5 = 7.5
   • y2=15⋅0.866≈12.99y_2 = 15 \cdot 0.866 \approx 12.99

Step 2: Add the components

Now, combine the x and y components of both movements:

• Total x-component: xtotal=x1+x2=0+7.5=7.5x_{\text{total}} = x_1 + x_2 = 0 + 7.5 = 7.5
• Total y-component: ytotal=y1+y2=20+12.99=32.99y_{\text{total}} = y_1 + y_2 = 20 + 12.99 = 32.99

Step 3: Calculate the resultant displacement

The resultant displacement RR is the vector sum of the total x and y components. We can find it using the Pythagorean theorem: R=xtotal2+ytotal2R = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2}

Substitute the values: R=(7.5)2+(32.99)2≈56.25+1084.60≈1140.85≈33.74 metersR = \sqrt{(7.5)^2 + (32.99)^2} \approx \sqrt{56.25 + 1084.60} \approx \sqrt{1140.85} \approx 33.74 \text{ meters}

Step 4: Find the direction of the resultant displacement

The direction (angle θ\theta) of the resultant displacement relative to north is given by: θ=tan⁡−1(xtotalytotal)\theta = \tan^{-1}\left(\frac{x_{\text{total}}}{y_{\text{total}}}\right)

Substitute the values: θ=tan⁡−1(7.532.99)≈tan⁡−1(0.227)≈12.8∘\theta = \tan^{-1}\left(\frac{7.5}{32.99}\right) \approx \tan^{-1}(0.227) \approx 12.8^\circ

So the resultant displacement is approximately 33.74 meters at an angle of 12.8° east of north.