Question

A puck moves 2.35 m/s in a -22.0° direction. A hockey stick pushes it for 0.215 s, changing ts velocity to 6.42 m/s in a 50.0° direction. What was the direction of the acceleration? ​

Answer 1

Approximately
`71.4^(\circ)`.

Step-by-step explanation:

Assume that all angles in this question are measured with respect to the positive
`x`-direction.

Given the magnitude and direction of both initial and final velocity, the direction of acceleration can be found in the following steps:

• Express initial and final velocity as vectors.
• Express acceleration as a vector.
• Take the inverse tangent (
  `\arctan`) of the ratio between the
  `y`-component and the
  `x`-component of the acceleration vector to find the angle between this vector and the
  `x\!`-axis. Verify if this angle is between the vector and the positive or the negative
  `\!x`-axis.

Let
`u` denote the initial velocity, and let
`\theta_(u) = (-22.0)^(\circ)` denote the direction of this velocity. The magnitude of initial velocity is
`\| u\| = 2.35\; {\rm m\cdot s^(-1)}`. Represent initial velocity as a vector:

`\displaystyle u = \begin{bmatrix}\|u\| \cos(\theta_(u))\\ \|u\| \sin(\theta_(u))\end{bmatrix}`.

Similarly, let
`v` denote the final velocity, and let
`\theta_(v) = 50.0^(\circ)` denote the direction of this velocity. The magnitude of final velocity is
`\| u\| =6.42\; {\rm m\cdot s^(-1)}`. Represent final velocity as a vector:

`\displaystyle v = \begin{bmatrix}\|v\| \cos(\theta_(v))\\ \|v\| \sin(\theta_(v))\end{bmatrix}`.

Acceleration
`a` is the rate of change in velocity. Let
`t` denote the duration of this motion (
`t \\e 0`.) Given the initial and final velocity of the motion, the acceleration of this motion would be:

`\begin{aligned}a &= (v - u)/(t) \\ &= (1)/(t)\, \left(\begin{bmatrix}\|v\| \cos(\theta_(v))\\ \|v\| \sin(\theta_(v))\end{bmatrix} - \begin{bmatrix}\|u\| \cos(\theta_(u))\\ \|u\| \sin(\theta_(u))\end{bmatrix}\right)\\ &= (1)/(t)\, \begin{bmatrix}\|v\| \cos(\theta_(v)) - \|u\| \cos(\theta_(u))\\ \|v\| \sin(\theta_(v)) - \|u\| \sin(\theta_(u))\end{bmatrix}\end{aligned}`.

The ratio between the
`y`- and
`x`-component of the acceleration vector is:

`\begin{aligned}(\|v\| \cos(\theta_(v)) - \|u\| \cos(\theta_(u)))/(\|v\| \sin(\theta_(v)) - \|u\| \sin(\theta_(u)))\end{aligned}`.

Apply the inverse tangent function (
`\arctan`) to this ratio to find angle between the acceleration vector and the positive
`x`-axis:

`\begin{aligned}\theta_(a) &= \arctan\left((\|v\| \cos(\theta_(v)) - \|u\| \cos(\theta_(u)))/(\|v\| \sin(\theta_(v)) - \|u\| \sin(\theta_(u)))\right) \\ &= \arctan\left((6.42\, \cos(50.0^(\circ)) - 2.35\, \cos(-22.0^(\circ)))/(6.42\, \sin(50.0^(\circ)) - 2.35\, \sin(-22.0^(\circ)))\right)\\ &\approx 71.4^(\circ)\end{aligned}`.

The output of
`\arctan` gives the angle between the vector and the
`x`-axis in the counterclockwise direction. However, this value does not distinguish between whether this angle is between the vector and the positive
`x`-axis or between the vector and the negative
`y`-axis. Verify the sign of the
`x`-component of the vector to determine the exact direction of the vector:

`\begin{aligned}& \|v\|\, \cos(\theta_(v)) - \|u\|\, \cos(\theta_(u)) \\ =\; & 6.42\, \cos(50.0^(\circ)) - 2.35\, \cos(-22.0^(\circ)) \\ \approx\; & 1.95\end{aligned}`.

Since the
`x`-component of this vector is positive, the output of
`\arctan` would be the angle between the vector and the positive
`x\!`-axis.