Chapter 4, Problem 1CQ

a. At this instant, is the particle in FIGURE Q4.1 speeding up, slowing down, or traveling at constant speed?
b. Is this particle curving to the right, curving to the left, or traveling straight?

To determine

The state of the particle: speeding up, slowing down or traveling at constant speed.

Answer to Problem 1CQ

Solution:

The particle is slowing down in the vertical axis and speeding up in the horizontal axis.

Explanation of Solution

The acceleration vector has a component that it is pointing in the opposite direction to velocity vector; this means the particle will slow down on this axis. The component in the horizontal axis will make the particle to speed up.

Given Info:

There is a constant speed vector at the vertical axis and a constant acceleration vector with a component at both axes.

Formula Used The particle has the next velocity vector equation:

  $velocity=\left(v_{x,o}+a_{x,0}t\right)\hat{x}+\left(v_{y,o}+a_{y,0}t\right)\hat{y}$

Where:

  $a_{x,0}=-a.\cos \theta$

It is negative because it is pointing to the left.

  $a_{y,0}=-a.\sin \theta$

It is negative because it is pointing downward.

  $v_{y,0}=v$

  $v_{x,0}=0$

Where a and v are the magnitudes of acceleration and velocity vector respectively.

Calculation

Substituting these values in the equation of velocity vector:

  $velocity=\left(-a.\cos \theta t\right)\hat{x}+\left(v-a.\sin \theta t\right)\hat{y}$

Conclusion

As we can see in the equation of velocity vector, the horizontal component is continuously increasing in the negative direction and the vertical component is slowing down since the initial velocity vector and the component of the acceleration vector are pointing in opposite directions.

To determine

The shape of the particle trajectory.

Answer to Problem 1CQ

Solution:

The particle curves to the left.

Explanation of Solution

Given Info:

There is a constant speed vector at the vertical axis and a constant acceleration vector with a component at both axes.

Every second, the velocity vector changes the position vector as well as the acceleration vector changes the velocity direction and magnitude. As we can see the acceleration vector is partially pointing to the left, therefore the velocity vector will be changing in that direction and so the position vector.

Formula We can obtain the position equation from the previous problem:

  $velocity=\left(-a.\cos \theta t\right)\hat{x}+\left(v-a.\sin \theta t\right)\hat{y}$

  $position=\int \left(-a.\cos \theta t\right)\hat{x}+\left(v-a.\sin \theta t\right)\hat{y} dt$

  $position=\left(-\frac{1}{2}a.\cos \theta t^2\right)\hat{x}+\left(vt-\frac{1}{2}a.\sin \theta t^2\right)\hat{y}$

Calculation

We can graph the last equation, and assuming the values:

  $\left|\overrightarrow{v}\right|=1\left[\frac{m}{s}\right], \left|\overrightarrow{a}\right|=0.5\left[\frac{m}{s^2}\right], \theta =40°$

  

Conclusion

As we can see, the particle trajectory curves to the left because the acceleration vector has a component on the horizontal axis which points the same way.