Examples

Section 2.3 Examples

Special Instruction: To tackle the physical problems the book uses ‘GUESS’ strategy for a systematic gatheing of information and finding a realistic and logistic solution. Here is what GUESS stands for.

G: Given, identify the given informations.
U: Unknown, identify the unknown that we are looking for.
E: Equation/s, set up the equation from given information.
S: Substitute, substitute the number/s in the equation/s.
S: Solve, solve the equation to obtain the answer.

Remark 2.3.1.

Before we start solving any problem, make sure all the physical quantities are taken in base unit of the systems of unit we are using. For example: mile per hour, mph is not a base unit in FPS sytem. so we need to convert mph into \(ft/s.\)

Example 2.3.2.

A car starts from rest and obtains a 20 \(m/s\) speed in 5 second. Find its acceleration.

Solution.

Given: initial velocity, \(v_i = 0 \,m/s\text{,}\) final velocity, \(v_f = 20\, m/s \text{,}\) time, \(t = 5 \,s\text{.}\)

Unknown: acceleration, \(a=\text{?}\)

Equation: we need to select the suitable equation from the list of equations of motion that contains initial velocity, final velocity, time, and acceleration of the object. Hence we are selecting equaion I here, that is \(v_{f}=v_{i}+at\text{.}\)

Substitute and Solve: substitute the value of given informations in equaion I, that is

\begin{align*} 20 \amp = 0 + a\times 5 \\ or, 20 \amp = 5a\\ \therefore a \amp = \frac{20}{5} = 4 \end{align*}

Answer.

Hence the acceleration of the car is 4 \(m/s^2.\) SI unit of acceleration is \(m/s^2\text{.}\)

Example 2.3.3.

A car starts from rest speeding up by 2 \(m/s^2\) acceleration for 1 minutes. Find the distance covered by the car.

Solution.

Given: \(v_i = 0 \,m/s\text{,}\) \(a = 2\, m/s^2 \text{,}\) \(t = 1 \times 60 = 60 \,s\text{.}\)

Unknown: distance \(d=\text{?}\)

Equation: \(d=v_{i}t+\frac{1}{2}at^2\text{.}\)

Substitute & Solve:

\begin{align*} d \amp = 0\times t + \frac{1}{2}at^2 \\ or, d \amp = 0+\frac{1}{2}\times 2\times 60^2 \\ \therefore d \amp = 3600 \end{align*}

Answer.

3600 m

Example 2.3.4.

A car is running with initial velocity 10 \(m/s\) acelerates to 2 \(m/s^2\) to reach at 50 \(m/s\text{.}\) Find the following:

How far did the car travell during this event.
How long did the car travell during this event.

Solution.

Given: \(v_i = 10 \,m/s\text{,}\) \(v_f = 50 \,m/s\text{,}\) \(a = 2 \,m/s^2 \text{.}\)

Unknown: \(d=\text{?}\)

Equation: \(v^2_{f}=v^2_{i}+2ad\text{.}\)

Substitute & Solve:

\begin{align*} 50^2 \amp = 10^2+ 2\times2\times d \\ or, 2500 \amp = 100+4d \\ or,4d \amp = 2500-100 \\ or, 4d \amp = 2500-100 \\ or, d \amp = \frac{2400}{4} = 600 \end{align*}

Answer.
\(\therefore d=600 \,m\)
Unknown: \(t=\text{?}\)

Equation: \(v_{f}=v_{i}+at\text{.}\)

Substitute & Solve:

\begin{align*} 50 \amp = 10+ 2\times t \\ or, 2 t \amp = 50-10 \\ or,2t \amp = 40 \\ or, t \amp = \frac{40}{2} = 20 \end{align*}

Answer.
\(\therefore t=20 \,s\)

Example 2.3.5.

How far you can go in 5 minutes if you are running at 10 \(km/h\text{?}\)

Solution.

Hint.

First convert the velocity into \(km/h\) to \(m/s\) and also convert time from minute to second before using equation of motion.

Given: \(v_i=v_f=v=10 km/h = 10\cdot\frac{1000m}{3600s} = 2.78 m/s, \quad t= 5\cdot60s=300s\)

\begin{align*} d \amp =vt+\frac{1}{2}at^2\\ d \amp =2.78 \times 300+\frac{1}{2}\times 0\times t^2\\ \therefore d \amp = 833 \,m \end{align*}

Example 2.3.6.

Which of the following sets of displacements might be able to return a car to its starting point?

4, 6, 8, and 15 km
2, 8, 10, and 25 km.
10, 30, 50, and 120 km
5, 20, 35, and 65 km.

Hint.

A car can return to its starting point, only when the sum of three displacements is greater than the value of the maximum displacement.

Answer.

Solution.

In case (a) sum of 4, 6, 8 =18 km which is greater than 15 km.

In case (b) sum of 2, 8, 10 =20 km which is smaller than 25 km and so on.

Example 2.3.7.

A ship travels 25 km to the east and then 50 km to the south. Find the ship’s displacement from its starting point.

Hint.

Draw a diagram to see all the displacements of the ship and then use pythagorous theorem to solve the final displacement.

Solution.

empty

Example 2.3.8.

A stone is thrown vertically up with velocity 20 \(m/s\text{.}\) Find the following:

Maximum height attains be the stone.
Total time of flight, \(T\)

Solution.

Given: \(v_i = 20 \,m/s\text{,}\) \(v_f = 0 \,m/s\text{,}\) \(g = -9.8 \,m/s^2 \text{.}\)

Unknown: \(h=\text{?}\)

Equation: \(v^2_{f}=v^2_{i}+2gh\text{.}\)

Substitute & Solve:

\begin{align*} 0 \amp = 20^2 - 2\times9.8\times h\\ or, 0 \amp = 400 -19.6 h \\ or, -400 \amp = -19.6 h\\ or, h \amp = \frac{400}{19.6} = 20.4 \end{align*}

Answer.
\(\therefore h=20.4 \,m\)
Unknown: \(t=\text{?}\)

Equation: \(v_{f}=v_{i}+at\text{.}\)

Substitute & Solve:

\begin{align*} 0 \amp = 20-9.8\times t \\ or, -20 \amp = -9.8 t \\ or, t \amp = \frac{20}{9.8}= 2 \end{align*}

Answer.
\(\therefore T =2t=2\times 2=4 \,s\)

You may select equation \(h=v_it+\frac{1}{2}gt^2\) and consider h=0, when stone comes back to the point from where it was thrown then \(h=0\) and \(T=2t\text{.}\)

\begin{align*} Now, 0 \amp = 20\times T-\frac{1}{2}\times 9.8\times T^2\\ or, 0 \amp = 20\times T-4.9\times T^2\\ or, 0 \amp = T(4.9T-20) \end{align*}

Answer.

\(\therefore T =20/4.9=4 \,s.\) After discarding \(T=0\text{.}\)

Example 2.3.9.

A stone dropped from a cliff reaches the ground in 4 s. Find the velocity at which it hits the ground.

Hint.

Think about stone’s initial velocity, use acceleration due to gravity to find final velocity.

Example 2.3.10.

A stone has been thrown upward with velocity 50 m/s.

How long does it take to come back on thrower’s hand (find the time of flight)?
What is its velocity then?
How high was it gone?

Example 2.3.11.

What is vertical motion?

Motion that occurs horizontally
Motion that occurs in a straight line
Motion that occurs in a vertical direction
Motion that occurs in a circular path

Answer.

Example 2.3.12.

What is free fall?

The motion of an object when it is moving at a constant speed
The motion of an object when it is not affected by gravity
The motion of an object when it is accelerating due to the force of gravity
The motion of an object when it is in a vacuum

Answer.

Example 2.3.13.

What is the formula for calculating the time of flight of a projectile?

\(\displaystyle t = v/f\)
\(\displaystyle t = d/v\)
\(\displaystyle t = 2v/g\)
\(\displaystyle t = 2d/g\)

Answer.

Example 2.3.14.

What is the difference between speed and velocity?

Speed is a scalar quantity and velocity is a vector quantity
Speed is a vector quantity and velocity is a scalar quantity
Speed and velocity are the same thing
Speed is measured in meters per second and velocity is measured in miles per hour

Answer.

Example 2.3.15.

A toy car is moving at a constant speed of 5 meters per second in a straight line. How far will it travel in 10 seconds?

Solution.

Using the formula \(d = vt\text{,}\) where d is the distance, v is the velocity, and t is the time, we can calculate the distance traveled by the toy car.

\begin{align*} d\amp = vt\\ \amp = 5 \,m/s \times 10 \,s = 50 \,meters \end{align*}

Therefore, the toy car will travel 50 meters in 10 seconds at a constant speed of 5 meters per second in a straight line.

Example 2.3.16.

A car is traveling at a constant speed of 20 meters per second. It takes the car 5 seconds to travel a distance of 100 meters. What is the distance the car will travel in 10 seconds at the same speed?

Solution.

We know that distance (d) is equal to speed (v) multiplied by time (t). Therefore, we can use the formula \(d = vt\) to solve this problem. First, we can calculate the distance traveled in 1 seconds, and then in 10 seconds.

\begin{align*} d \amp = vt\\ \amp = 20 \,m/s\times 1 \,s =20\,m \\ \therefore d \amp = 20\,m \times 10 \,s = 200 \,m \end{align*}

Therefore, the car will travel a distance of 200 meters in 10 seconds at the same speed.

Example 2.3.17.

A car travels on a straight road at a constant speed of 60 km/h. How far will the car travel in 2 hours?

Solution.

Since the car is traveling at a constant speed, we can use the formula:

\begin{equation*} d=vt \end{equation*}

where d is the total distance traveled, v is the constant speed of the car, and t is the total time the car travels. Plugging in the given values, we get:

\begin{align*} d\amp =vt\\ \amp = 60\,km/h\times 2h=120 \,m \end{align*}

Therefore, the car will travel a total distance of 120 kilometers in 2 hours at a constant speed of 60 km/h.

Example 2.3.18.

A ball is thrown vertically upward with an initial velocity of 20 m/s. What is the maximum height reached by the ball and how long does it take to reach this height? (Assume no air resistance and a gravitational acceleration of \(9.8 \,m/s²\))

Solution.

When an object is thrown vertically upward, it is subjected to a negative acceleration due to gravity, which causes it to slow down until it reaches its maximum height, after which it starts to fall back to the ground with a positive acceleration due to gravity.

To solve this problem, we can use the following kinematic equations for motion with constant acceleration:

\begin{align*} v_f \amp = v_i + at \\ d\amp = v_it + 1/2a*t^2\\ v_f^2 \amp = v_i^2 + 2ad \end{align*}

where \(v_i\) and \(v_f\) are the initial and final velocities, \(a\) is the acceleration, \(d\) is the displacement, and \(t\) is the time. At the maximum height, the velocity of the ball is zero, so we can use the first equation to find the time it takes to reach this height:

\begin{align*} v_f \amp = v_i + at \\ 0 \amp = 20 \,m/s - 9.8 \,m/s²\times t \\ t \amp = 20/9.8 \,seconds ≈ 2.04 \,seconds \end{align*}

Now we can use the second equation to find the maximum height:

\begin{align*} d \amp = v_it + 1/2a\times t^2 \\ d \amp = 20 \,m/s \times 2.04 \,s + 1/2 \times (-9.8\,m/s²) \times (2.04 \,s)^2 \\ d \amp = 20.4 \,m \end{align*}

Therefore, the maximum height reached by the ball is 20.4 meters, and it takes approximately 2.04 seconds to reach this height.

Example 2.3.19.

A car is traveling at \(20 \,m/s\) and accelerates at a rate of \(4 \,m/s^2\) for 8 seconds. What is the final velocity of the car?

Solution.

We can use the formula for accelerated motion, which relates the initial velocity, final velocity, acceleration, and time:

\begin{align*} v_f\amp =v_i +at \\ v_f\amp =20 \,m/s + 4\,m/s^2 \times 8\,s \\ \amp = 20 \,m/s+32\,m/s = 52\,m/s \end{align*}

Therefore, the final velocity of the car after accelerating at a rate of \(4 \,m/s^2\) for 8 seconds is \(52 \,m/s\text{.}\)

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