Exercises On Uniform Linear Motion

Since the airplane flies directly in a straight line from A to B, and maintains its velocity constant, the equation of uniform rectilinear motion is applicable in this case:

$$ x(t)=x_0 + v\cdot t$$

The initial position is $latex x_0 =0$, the speed of the plane is $latex v=830\hspace{1 mm}\frac{km}{h}$ and the time is $latex t=210\hspace{1mm} minutes$. Before substituting the values ​​in the formula, it is necessary to express the time in hours, since the distance that separates the cities in kilometers is requested. Therefore:

$$210\hspace {1 mm}minutes=210\times \frac{1\hspace{1mm} hours}{60\hspace{1 mm} minutes}=3.5\hspace{1 mm}hours$$

Then:

$$ x(t)=v\cdot t$$

$$ x(3.5)=830\hspace{1 mm}\frac{km}{h}\times 3.5\hspace{1 mm}h=2905\hspace{1mm}km$$

The distance between cities A and B is $latex 2905\hspace{1mm}km$.

We will assume, for simplicity, that the motion of the continents is uniform rectilinear. Now, we can use the equation:

$$ x(t)=x_0 + v\cdot t$$

And we want:

$$\Delta x = x-x_0$$

Whose value, according to the statement, is:

$$\Delta x = x-x_0=500\hspace{1mm}km$$

$$\Delta x = v\cdot t$$

Solving for the time $latex t$:

$$t=\frac{\Delta x}{v}$$

Before substituting, we need to verify that the units are homogeneous. Kilometers can be expressed in centimeters, or centimeters in kilometers, using scientific notation.

In any case, the time will be in years:

$$t=\frac{\Delta x}{v}=\frac{5\times 10^7\hspace{1mm}cm}{3\hspace{1mm}\dfrac{cm}{year}}=1.7\times 10^7\hspace{1mm}years$$

One million years is $latex 1\times 10^6\hspace{1mm}years$, so the continents will have separated 500 km after about 17 million years.

Starting from the equation:

$$ x(t)=x_0 + v\cdot t$$

The values ​​$latex x_0= 4\hspace{1mm}m$, $latex v = -10\hspace{1mm}m/s$ and $latex t= 5\hspace{1mm} s$ are taken and these values ​​are substituted to obtain the equation for the position.

The negative sign in the velocity is due to the fact that the object is moving in a negative direction, according to the statement:

$$ x(t)=4 -10\cdot t$$

After t=5 s, the object will be in:

$$ x(5)=4\hspace{1mm}m - \left(10\hspace{1mm}\frac{m}{s}\times5\hspace{1mm} s\right)=-46\hspace{1mm}m$$

Therefore, after t=5 s, the object will be in:

$$ x(5)=-46\hspace{1mm}m$$

Starting from the equation for uniform rectilinear motion, solve for the time:

$$ x(t)=v\cdot t\Rightarrow t=\frac{x}{v}$$

Before substituting numerical values ​​in the statement, we need to make sure that the units match, so we can use the speed of light as $latex c=300,000\hspace{1 mm}km/s$, so that time calculated is in seconds:

$$ t=\frac{x}{v}=\frac{150000000\hspace{1 mm}km}{300000\hspace{1mm}\dfrac{km}{s}}=500 \hspace{1mm}s=8.33\hspace{1 mm} minutos$$

The light we see from the Sun took approximately 8 minutes to reach the Earth.

a) According to the information in the statement, the initial position of the particle is $latex x_0=4\hspace{1mm}m$, and since we know its speed, when substituting values ​​in the equation of motion, it results:

$$x(t)=4+8t$$

b) To calculate the time, we substitute $latex x(t)=12\hspace{1mm}m$ and solve for the time:

$$t=\dfrac{12\hspace{1mm}m-4\hspace{1mm}m}{8\hspace{1mm}m/s}=1\hspace{1mm}s$$

c) After 5 seconds, the particle will be at the position $latex x(5)$, given by:

$$x(5)=4\hspace{1mm}m+\left(8\hspace{1mm}\frac{m}{s}\times 5\hspace{1mm}s\right)=44\hspace{1mm}m$$

d) The displacement in $latex t=5\hspace{1mm}s$ is:

$$\Delta x= x_{final} -x_{initial} =44\hspace{1mm}m- 4\hspace{1mm}m=40\hspace{1mm}m$$

a) The first step is to convert from minutes to hours:

$$120\hspace{1 mm}minutes=2\hspace{1 mm}hours$$

Since the starting point is the same, $latex x_0=0$ is taken for the two cars.

We will call car 1 the slowest and car 2 the fastest. In that case, car 2 overtakes car 1:

$$x_1=75\frac{km}{h}\times2\hspace{1 mm}hours=150\hspace{1 mm}km$$

$$x_2=90\frac{km}{h}\times2\hspace{1 mm}hours=180\hspace{1 mm}km$$

Therefore, after $latex t=2\hspace{1mm} hours$, the cars are separated by a distance of:

$$ (180 - 150) \hspace{1 mm}km=30 \hspace{1 mm}km$$

b) In this case, to distinguish the directions in which the cars move, it is enough to assign a negative speed to one of them. For example, you can make $latex v_2=-90\hspace{1mm}\frac{km}{h}$. This way:

$$x_1=75\frac{km}{h}\times2\hspace{1 mm}hours=150\hspace{1 mm}km$$

$$x_2=-90\frac{km}{h}\times2\hspace{1 mm}hours=-180\hspace{1 mm}km$$

After $latex t=2\hspace{1mm} hours$, the cars, which have been moving away from each other since the beginning, will be separated by a distance of:

$$ [150-(-180)] \hspace{1 mm}km=330 \hspace{1 mm}km$$

a) The car is initially 125 m from the origin:

$$x_0=125\hspace{1mm}m$$

b) The car is stopped at the following time intervals:

• From $latex t=0$ to $latex t = 2\hspace{1 mm} s$
• From $latex t=4$ to $latex t = 6\hspace{1 mm} s$
• From $latex t=7$ to $latex t = 2\hspace{1 mm} s$

c) The car has negative velocity in the interval:

• From $latex t=6$ to $latex t = 7\hspace{1 mm} s$

d) The car has positive velocity in the interval:

• From $latex t=2$ to $latex t = 4\hspace{1 mm} s$
• From $latex t=8$ to $latex t = 10\hspace{1 mm} s$

e) The total distance is calculated by adding the distances that the car travels in the corresponding time interval:

• From $latex t=0$ to $latex t = 2\hspace{1 mm} s$ it's at rest, so, $latex \Delta x_1=0\hspace{1mm}m$
• From $latex t=2$ to $latex t = 4\hspace{1 mm} s$ moves $latex \Delta x_2=225-125\hspace{1mm}m=100\hspace{1 mm}m$
• From $latex t=4$ to $latex t = 6\hspace{1 mm} s$ it's at rest, so, $latex \Delta x_3=0\hspace{1mm}m$
• From $latex t=6$ to $latex t = 7\hspace{1 mm} s$ the car returns to the origin, $latex \Delta x_4=225\hspace{1mm}m$
• From $latex t=7$ to $latex t = 8\hspace{1 mm} s$ it's at rest, so, $latex \Delta x_5=0\hspace{1mm}m$
• From $latex t=8$ to $latex t = 10\hspace{1 mm} s$ moves $latex \Delta x_6=37.5\hspace{1mm}m$

$$Total ~Distance\hspace{1mm} = 100\hspace{1 mm}m+225\hspace{1mm}m+37.5\hspace{1mm}m=362.5\hspace{1mm}m$$

f) The movement consists of six different sections, in sections 1, 3 and 5 it is at rest, in sections 2 and 6 it has positive speed and in section 4 it has negative speed. The speed is equivalent to the slope of the straight line in each section:

• $latex v_1=0\hspace{1mm}m/s$
• $latex v_2=\dfrac{225m-125m}{4s-2s}=+50\hspace{1mm}m/s$
• $latex v_3=0\hspace{1mm}m/s$
• $latex v_4=\dfrac{225m-0m}{6s-7s}=-225\hspace{1mm}m/s$
• $latex v_5=0\hspace{1mm}m/s$
• $latex v_6=\dfrac{37.5m-0m}{10s-8s}=+18.75\hspace{1mm}m/s$

g) The average speed is the quotient between the total distance traveled and the time the movement lasts:

$$v_m=\dfrac{362.5\hspace{1mm}m}{10\hspace{1mm}s}=36.25\hspace{1mm}m/s$$

a) Let the initial time be $latex t_0 =0$. At that moment, object 1 is assigned the position $latex x_1=0$ and object 2 is assigned the position $latex x_2 =150\hspace{1mm}m$. Their respective speeds are: $latex v_1=+6\hspace{1mm}m/s$ and $latex v_2=-9\hspace{1mm}m/s$, therefore:

$$x_1(t)=6t$$

$$x_2(t)=150-9t$$

In order to determine the time when they run into each other, the respective equations are equated:

$$6t=150-9t$$

$$15t=150$$

$$t=10\hspace{1mm}s$$

The objects intersect after $latex t = 10\hspace{1mm}s$.

b) After 15 seconds from the initial instant, 5 seconds have passed after they crossed each other. In that time, object 1 traveled the following distance.

$$x_1 = 6t=6 \hspace{1 mm}\frac{m}{s}\times 5\hspace{1mm} s=30 \hspace{1mm}m$$

And the object traveled this other distance:

$$x_2=6t=9 \hspace{1 mm}\frac{m}{s}\times 5\hspace{1mm} s=45\hspace{1mm}m$$

Remember that the distance is always positive, only that the objects move in opposite directions, therefore, the distance that separates them 15 seconds after they have intersected is:

$$30\hspace{1mm}m+45\hspace{1mm}m =75 \hspace{1mm}m$$

Suppose the red car travels at 12 km/h and the green car travels at 18 km/h. Since they move perpendicularly, their straight trajectories form the sides of a right triangle, while the distance separating them is the hypotenuse.

From the Pythagorean theorem:

$$ (18t)^2+(12t)^2 = 900^2 \rightarrow 324t^2+144t^2=810000$$

$$468t^2=810000$$

$$t=\sqrt{\dfrac{810000}{468}}\hspace{1mm}s=41.6\hspace{1mm} s$$

After 41.6 s after the crossing, the cars will be 900 m apart.

a) The particle position function is a piecewise function with three sections.

Part 1

Initial position: $latex x_0=-8\hspace{1mm}m$

Duration: starts at $latex t=0$ and ends at $latex t=10\hspace{1mm}s$

Velocity: $latex + 2m/s$

Equation: $latex x(t)=-8+2t$

Final position: $latex x(10)=-8+2\times 10\hspace{1mm} m= 12 \hspace{1mm} m$.

Part 2

Initial position: $latex x(10)=12\hspace{1mm}m$

Duration: starts at $latex t=10\hspace{1mm}s$ and ends at $latex t=16\hspace{1mm}s$

Speed: null

Equation: $latex x(t)= 12\hspace{1mm}m$ (The particle remained at rest)

Final position: $latex x(16)=12\hspace{1mm}m$.

Part 3

Initial position: $latex x(16)=12\hspace{1mm}m$

Duration: starts at $latex t=16\hspace{1mm}s$ and ends at $latex t=25\hspace{1mm}s$

Velocity: $latex v=-5\hspace{1mm}m/s$

Equation: $latex x(t)=12-5(t-16)$ (The time is counted from t=0 and this section began 16 seconds after the movement began, so we must subtract 16 from the variable).

Final position: $latex x(25)=12-5\times (25-16)\hspace{1mm}=-33\hspace{1mm}m$

Putting the three expressions together:

$$x(t) = \left\{\begin{array}{rl}  -8+2t & \text{if } 0 \leq t\leq10\\   12 & \text{if } 10 \leq t\leq16\\ 12-5(t-16) & \text{if } 16\leq t\leq25\end{array} \right. $$

b) Graph of position as a function of time:

c) At $latex t=18\hspace{1mm}s$ the particle is in $latex x=2\hspace{1mm}m$.

d) At the end of the movement, the particle was in the position $latex x=-33\hspace{1mm}m$.

e) Graph of velocity as a function of time:

f) The distance traveled during the first 20 seconds is numerically equivalent to the area between the graph and the horizontal axis.

As they are rectangles, the area is calculated by multiplying base by height:

$$d_{20}= |10\hspace{1mm}\times2\hspace{1mm}m/s|+|{4\hspace{1mm}\times(-5\hspace{1mm}m/s)}|=40\hspace{1mm}m$$

Note that the area is always a positive quantity.

g) The total distance traveled by the object is numerically equivalent to the area highlighted in color in the figure.

You can use the result of the previous item and add the missing area, which is the area of the rectangle whose base goes from 20s to 25s:

$$d_{25}=40\hspace{1mm}m +|{5\hspace{1mm}s\hspace{1mm}\times(-5\hspace{1mm}m/s)}|=65\hspace{1mm}m$$.

h) The average speed of the trip is calculated by dividing the total distance traveled by the time taken:

$$v_m=\dfrac{65\hspace{1mm}m}{10\hspace{1mm}s}=6.5\hspace{1mm}m/s$$