Understanding Motion in One Dimension

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Marcus throws a ball directly up in the air with an initial velocity of $\small 35\frac{m}{s}$ . How high will the ball go?

$\small g=-9.8\frac{m}{s^2}$

Explanation

Remember that the final velocity at an object's highest point is equal to zero. We can use the following equation to solve for our height ( $\Delta{y}}$ ), using the initial velocity, final velocity, and acceleration due to gravity.

$V_{f}^{2}=V_{i}^{2}+2a\Delta{y}}$

$(0\frac{m}{s})^2=({35\frac{m}{s}})^{2}+2(-9.8\frac{m}{s^2})\Delta{y}}$

$0=1225\frac{m^2}{s^2}+(-19.6\frac{m}{s^2})\Delta{y}}$

$-1225\frac{m^2}{s^2}=(-19.6\frac{m}{s^2})\Delta{y}}$

$\frac{-1225\frac{m^2}{s^2}}{-19.6\frac{m}{s^2}} =\Delta{y}}$

$62.5m=\Delta{y}$

Suppose a recreational biker averages $8.05\frac{m}{s}$ on a twenty-mile ride, equal to . A professional biker has an average speed of $11.1\frac{m}{s}$ . The professional happens to be riding on the same path, but started behind the recreational biker. The two are both headed for the same destination. Who would reach the end of the path first, and how far behind would the other biker be?

The professional finishes first and the other biker is behind

Both bikers reach destination at same time

The recreational finishes first and the other biker is behind

Explanation

To solve this problem we can simply examine each biker separately and see how long it would take them to reach the destination. Let's begin with the recreational biker. The recreational biker needs to ride at $8.05\frac{m}{s}$ . Using the definition of velocity, we can find his final time.

$v=\frac{d}{t}$

$t=\frac{d}{v}$

$t=\frac{3.22*10^4m}{8.05\frac{m}{s}}$

The professional has a distance of , plus the that he's behind.

We know that the velocity of the professional biker is $11.1\frac{m}{s}$ . Using this velocity and his total distance, we can find the time that it takes him to reach the end of the path.

$v=\frac{d}{t}$

$t=\frac{d}{v}$

$t=\frac{4.42*10^4m}{11.1\frac{m}{s}}$

The time of the professional biker is less than that of the recreational biker, meaning that the professional will finish first. Now we need to find the distance between the two bikers at this point. Use the recreational biker's velocity and the time difference between the two bikers to solve for the distance that the recreational biker has left on the path.

The recreational biker will ride for at $8.05\frac{m}{s}$ to finish the path.

$v=\frac{d}{t}$

$d=(8.05\frac{m}{s})(18s)$

$d=144.9m\approx145m$

A tennis ball is thrown straight up and it is caught at the same height the person released the ball from their hand. Which of the following is false? Ignore air resistance.

Acceleration and velocity point in the same direction the entire time

The time the tennis ball is traveling up is equal to the time it falls down

The speed of the tennis ball as it leaves the person's hand is the same as when it is caught

All of these answers are true

The velocity will change sign at the the top of the motion

Explanation

The question asks to point out the false statement. Everything on Earth is accelerated downwards by gravity, all the time, by $9.81\frac{m}{s^2}$ . Think of gravity as having a negative sign. When the ball is thrown up, acceleration is working against the velocity slowing the ball down. Their signs are opposite. But when the ball is falling back down to Earth, the velocity and acceleration have the same sign. So velocity and acceleration will not always have the same sign.

If gravity could be said to have a negative sign since it pulls everything downward, then an upward velocity would have a positive (and opposing sign). At the top of the trajectory when the ball's upward velocity is finally overcome by gravity, the sign of the velocity becomes negative as it now points back down to Earth. So it is true velocity changes sign at the top.

Since the ball is caught at the same height, both the time up and time down are equal and it will be traveling at the same speed. Since gravity is the only force acting on it, the ball loses all its velocity on the way up and regains that exact amount by the time it reaches the height it started the journey. This also makes the time up equal to the time it falls. If the same force acts with same strength (gravity) the entire time, why would either of these change?

Laura throws a ball vertically. She notices it reaches a maximum height of 10 meters. What was the initial velocity of the ball?

$\small g=-9.8\frac{m}{s^2}$

$14\frac{m}{s}$

$196\frac{m}{s}$

$144\frac{m}{s}$

$29\frac{m}{s}$

$22.2\frac{m}{s}$

Explanation

Remember that at the highest point, the velocity in the y-direction is equal to zero. Using the given values and the equation below, we can solve for the initial velocity.

$V_f^2=V_i^2+2a\Delta y$

$(0\frac{m}{s})^2=V_i^2+2(-9.8\frac{m}{s^2})(10m)$

$0\frac{m^2}{s^2}=V_i^2-196\frac{m^2}{s^2}$

$196\frac{m^2}{s^2}=V_i^2$

$\sqrt{196\frac{m^2}{s^2}}=\sqrt{V_i^2}$

$14\frac{m}{s}=V_i$

Two balls, one with mass and one with mass , are dropped from above the ground. How long does it take the ball to hit the ground?

$\small g=-9.8\frac{m}{s^2}$

Explanation

The equation for the vertical motion for an object in freefall is:

$\Delta y=\frac{1}{2}at^2$

Notice that there is no place for mass anywhere in this equation. This means that the two balls will be in the air for the same amount of time. We simply need to use the distance and acceleration to solve for the time.

Remember that even though the height is , the DISPLACEMENT will be . Displacement is a vector; since the direction of the distance is downward, the displacement will be negative.

$-30m=\frac{1}{2}(-9.8\frac{m}{s^2 })t^2$

$\frac{-30m}{-9.8\frac{m}{s^2}}=\frac{1}{2}t^2$

$3.06s^2=\frac{1}{2}t^2$

$\sqrt{6.12s^2}=t^2$

A ball rolls along a table with a constant velocity of $0.8\frac{m}{s}$ . If it rolls for , how long was it rolling?

Explanation

The relationship between constant velocity, distance, and time can best be illustrated as $\Delta x =v*t$ .

We are given the velocity and the distance, allowing us to solve for the time that the ball was in motion.

$\Delta x =v*t$

$10m =0.8\frac{m}{s}*t$

Isolate the variable for time.

$\frac{10m}{0.8\frac{m}{s}} =t$

If you threw a tomato upwards with an initial velocity of $12\frac{m}{s}$ , at what time (in seconds) would the tomato hit the ground?

$g=10\frac{m}{s^2}$

Explanation

Gravity accelerates everything downward by $10\frac{m}{s^2}$ . When the tomato is thrown upward with some velocity, gravity immediately begins to slowly reduce this velocity since the acceleration opposes the direction of the velocity.

$V_{0}=12\frac{m}{s}$ (initial velocity)

$V=0\frac{m}{s}$ (final velocity is equal to 0)

$t=\frac{V-V_{0}}{a}=\frac{0-12\frac{m}{s}}{-10\frac{m}{s^2}}=1.2s$

In 1.2s the tomato will reach its maximum height. If you have ever thrown a ball upward you may have noticed how it appears to stop at the peak. We have just calculated the time it takes for that ball to appear to stop for a very small time and fall back down. Since our tomato must travel back to Earth, we double the time for its up and down motion (since they equal each other) to get 2.4s as the final answer.

A crate slides along a frictionless surface. If it maintains a constant velocity of $3\frac{m}{s}$ , what is the net acceleration on the crate?

$0\frac{m}{s^2}$

$3\frac{m}{s^2}$

$60\frac{m}{s^2}$

$6.67\frac{m}{s^2}$

$20\frac{m}{s^2}$

Explanation

Acceleration is the change in velocity per unit time.

$a=\frac{v_2-v_1}{t}$

Since the velocity does not change from one moment to the next, then there must be no net acceleration on the object.

$a=\frac{3\frac{m}{s}-3\frac{m}{s}}{t}=\frac{0\frac{m}{s}}{t}$

$a=0\frac{m}{s^2}$

Jenny throws a ball directly up in the air. She notices that it changes direction after approximately 3 seconds. What was the initial velocity of the ball?

$\small g=-9.8\frac{m}{s^2}$

$29.4\frac{m}{s}$

$35.0\frac{m}{s}$

$19.6\frac{m}{s}$

$8.8\frac{m}{s}$

$9.8\frac{m}{s}$

Explanation

The ball will travel upwards to the highest point, then change direction and travel downwards. Remember that the velocity in at the highest point is equal to zero. We can use the following equation and our known data to solve for the initial velocity.

$V_{f}=V_{i}+at$

$0\frac{m}{s}=V_{i}+(-9.8\frac{m}{s^2})(3s)$

$0\frac{m}{s}=V_{i}-29.4\frac{m}{s}$

$29.4\frac{m}{s}=V_{i}$

A ball is thrown vertically with a velocity of . What is its velocity at the highest point in the throw?

$0\frac{m}{s}$

$\frac{v}{2}$

There is insufficient information to solve

Explanation

When examining vertical motion, the vertical velocity will always be zero at the highest point. At this point, the acceleration from gravity is working to change the motion of the ball from positive (upward) to negative (downward). This change is represented by the x-axis on a velocity versus time graph. As the ball changes direction, its velocity crosses the x-axis, momentarily becoming zero.

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