Question

1. a snowboarder carves down a black diamond run accelerating 4.1 m/s^2 n. then the snowboarder takes a bank and accelerates at 2.7 m/s^2 ne. what is the overall acceleration of the snowboarder’s motion up to this point?

Explanation:

Step1: Resolve the second acceleration into components

The second acceleration \(a_2 = 2.7\ m/s^2\) in the northeast (NE) direction. The angle \(\theta = 45^{\circ}\) for NE.
The \(x\) - component \(a_{2x}=a_2\cos45^{\circ}\) and the \(y\) - component \(a_{2y}=a_2\sin45^{\circ}\).

$$a_{2x}=2.7\times\frac{\sqrt{2}}{2}\approx2.7\times0.707 = 1.91\ m/s^2$$

$$a_{2y}=2.7\times\frac{\sqrt{2}}{2}\approx1.91\ m/s^2$$

The first acceleration \(a_1 = 4.1\ m/s^2\) in the \(y\) - direction (assuming the initial direction is \(y\) - axis).

Step2: Find the total \(x\) and \(y\) accelerations

The total \(x\) - acceleration \(a_x=a_{2x}\approx1.91\ m/s^2\)
The total \(y\) - acceleration \(a_y=a_1 + a_{2y}\)

$$a_y=4.1+1.91=6.01\ m/s^2$$

Step3: Calculate the magnitude of the resultant acceleration

Using the Pythagorean theorem \(a=\sqrt{a_x^{2}+a_y^{2}}\)

$$a=\sqrt{(1.91)^{2}+(6.01)^{2}}=\sqrt{3.65+36.12}=\sqrt{39.77}\approx 6.3\ m/s^2$$

Step4: Calculate the direction of the resultant acceleration

Using the formula \(\tan\theta=\frac{a_y}{a_x}\)

$$\tan\theta=\frac{6.01}{1.91}\approx3.15$$

\(\theta=\arctan(3.15)\approx72.4^{\circ}\) above the \(x\) - axis (or \(17.6^{\circ}\) east of north)

Answer:

The overall acceleration is approximately \(6.3\ m/s^2\) at an angle of about \(72.4^{\circ}\) above the \(x\) - axis (or \(17.6^{\circ}\) east of north)