It is a straight forward 2 dimensional kinematics problem.

A position vector that is a function of time is given, hence we can get the velocity vector by differentiating the position function.

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# Physics Problems

## General two dimensional kinematics problem

## Relative Velocity Problem

## Circular motion with constant acceleration problem

## Projectile Problem

## Projectile Problem

Select Solved Problems for practice

From Young and Freedman’s University physics with modern physics textbook

It is a straight forward 2 dimensional kinematics problem.

A position vector that is a function of time is given, hence we can get the velocity vector by differentiating the position function.

From Serway and Jewett’s Physics for scientists and engineers textbook (6th edition)

This problem from Serway and Jewett’s Physics for scientists and engineers with modern physics textbook (9th edition)

Let’s analyze the wording of this problem:

- A ball swings in a vertical circle… then it is a circular motion problem.
- A given acceleration at specific position in space which is 36.9° past the lowest point on way up for the ball..then the given acceleration does not seem not to be a radial acceleration (see the diagram below in solution).

The information given in this problem:

- Radius of circular motion is 1.5 m (R=1.5m).
- The acceleration of the ball is a=-22.5iˆ+20.2jˆ m/s² at 36.9° away from the vertical axis (y-axis).

What the problem need:

- Sketch the vector diagram showing the components of the acceleration.
- Find the magnitude of radial acceleration.
- Find the speed and velocity of the ball.

SOLUTION:

**Sketch the vector diagram showing the components of the acceleration**.

The diagram of acceleration vector

2. **Find the magnitude of radial acceleration.**

The magnitude of radial acceleration could be found by using the horizontal and vertical components of the given acceleration:

3. **Find the speed and velocity of the ball.**

The speed could be found using the radial acceleration and the radius of the circular motion:

Problem from Young and Freedman’s University Physics with Modern Physics textbook

First we need to analyze the wording of the problem:

- A snow ball rolls off a barn roof.. that mean it is a projectile problem
- The height is y1=14 m and y2=0 m (the ground) then Δy=-14 m
- The range is x2= 4 m and x1=0 m (the barn side has the x-axis origin).

What are the given information:

- Angle at which the ball is leaving the roof is 40°.
- The initial velocity (vo) =7 m/s
- Δy=-14 m
- Δx=4 m

What the problem need:

- How far the ball will fall away from the barn (Δx=?)
- Draw x-t, y-t, vx-t, and vy-t graphs for the motion in part (1)
- If a man (his height= 1.9m) is standing at Δx=4 m, do the ball hit him when it falls?

SOLUTION:

**How far the ball will fall away from the barn (Δx=?)**

2.** Draw x-t, y-t, vx-t, and vy-t graphs for the motion in part (1)**

3.** If a man (his height= 1.9m) is standing at Δx=4 m, do the ball hit him when it falls?**

For this question, it is easy to find the time when Δx=4 m:

While the man height is 1.9 m then the ball will be far from hitting him and he will be fine.

The following problem is from Young and Freedman’s University Physics with modern physics:

Let’s analyze some of the wording of the problem:

- A rocket is fired at an angle from the top of a tower…that means it is a projectile problem.
- The height of tower h
_{o}= 49.3 m and the origin of coordinates is at the base of the tower.. then Δy=y_{2}-y_{1}=0-49.3= -49.3 m

Now what are the given information:

- h
_{0}= 49.3 m or we can say Δy= -49.3 m - Position coordinates are x(t) and y(t) given by x(t)=A+Bt
^{2}and y(t)=C+Dt^{3}. We can say that at t=0 s, x(0)=A and y(0)=C. When it is a tower and the base of the tower is the origin of the coordinates then x(0)=0 m and y(0)=49.3 m. - Acceleration of the rocket at time t= 1.5 s is a=5.4i+3.2j m/s
^{2}

What the problem needs:

- What are A,B,C, and D constants?
- What are the acceleration vector (a=?) and velocity vector (v=?) at the moment it got fired?
- What are the x- and y- component of velocity at time t=17.1 s after firing the rocket?
- What are the x- and y-component of position at time t=17.1s after firing the rocket?

SOLUTION:

**What are A,B,C, and D constants?**

At first we can see that when x(0)=0 m = A then A=0 m and when y(0)=49.3 m = C then C= 49.3 m. We can use the the coordinate function x(t) and y(t) to get v(t) and a(t).

2. **What are the acceleration vector (a=?) and velocity vector (v=?) at the moment it got fired?**

Velocity and acceleration vectors already written above, then velocity and acceleration at the moment of firing the rocket are:

3. **What are the x- and y- component of velocity at time t=17.1 s after firing the rocket?**

4. **What are the x- and y-component of position at time t=17.1s after firing the rocket?**