Motion in One Dimension - AP Physics 1

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Question

If a 15kg ball takes five seconds to strike the ground when released from rest, at what height was the ball dropped?

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Answer

Using the equation we can find the distance at which the ball was dropped. Notice that the mass of the ball does not matter in this problem. We are told that the ball is dropped from rest making, , thus we have . When we plug in our values, and assuming that acceleration is equal to gravity (10m/s2) we find that = 125m.

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Question

How far will an object travel after ten seconds if it is dropped into a bottomless pit?

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Answer

Since the object is dropped, the inital velocity is zero. Gravity is the only acceleration, the time is ten seconds, and the distance at which the object travels is unknown.

The equation can be used to find the distance traveled.

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Question

How long does it take an object to travel a distance of 30m from rest at a constant acceleration of 2m/s2?

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Answer

Using the equation , we can solve for time.

Since the object started at rest, . Now we are left with the equation .

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

Plugging in the remaining values we can find that t = 5.5s.

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Question

You are driving at a speed of $20\frac{m}{s}$ and suddenly, a tree falls down on the road blocking your path. You slam on your brakes to avoid hitting the fallen tree and thus, come to a complete stop. You were at a distance of away from the tree when you hit the brakes. Assuming that your vehicle does not skid, what is the minimum deceleration needed to avoid hitting the fallen tree?

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Answer

Use the following kinematic equation where the initial velocity is , final velocity is , and the distance traveled is $\Delta x$ .

$v_f^2=v_o^2+2a\Delta x$

We can use the values in the question to solve for the acceleration.

$(0\frac{m}{s})^2=(20\frac{m}{s})^2+2a(28m)$

We rearrange the equation to solve for the acceleration.

$a=\frac{0^2-(20\frac{m}{s})^2}{2(28\ \text{m})}$

$a=-7.14\ \frac{\text{m}}{\text{s}^2}$

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Question

A person on top of a tall building drops a rock. How long will it take for the rock to reach the ground? Ignore air resistance.

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Answer

Use the following kinematic equation: $y=y_o+v_ot+\frac{1}{2}at^2$ .

We can choose the ground to be the zero distance so that and $y_o=200\ \text{m}$ .

Also, the initial speed is zero.

The kinematic equation simplifies using these values.

$0=y_o+\frac{1}{2}at^2$

Solve to isolate the time variable.

$t=\sqrt{\frac{-2y_o}{a}}$

We know that the acceleration is the acceleration due to gravity. Now we can plug in the known values and solve.

$t=\sqrt{\frac{-2(200\ \text{m})}{-9.8\ \frac{\text{m}}{\text{s}^2}}}$

$t=\sqrt{40.8s^2}=6.4s$

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Question

A block is placed at a height of on a ramp at an incline of . It slides down from rest with an acceleration of $2.2\frac{m}{s^2}$ due to a frictional force. What is the velocity of the block when it reaches the bottom of incline?

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Answer

To find the final velocity we use the following kinematic equation.

$v_f^2=v_o^2+2a\Delta x$

The block starts from rest, so . The incline is eighteen meters high, so we can find the distance (hypotenuse) as follows:

$\Delta x=\frac{18m}{sin24^\circ}$

$\Delta x = 44.25m$

We know that the acceleration is $a=2.2\frac{m}{s^2}$ .

Plug in the values to find the velocity at the bottom of the incline.

$v_f^2=(0\frac{m}{s})^2+2(2.2\frac{m}{s^2})(44.25m)$

$v_f=\sqrt{194.7\frac{m^2}{s^2}}=14\frac{m}{s}$

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Question

A ball is thrown horizontally off a cliff of height of $\small 50m$ with an initial velocity of $\small 15\frac{m}{s}$ . How far from the cliff will the ball land?

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Answer

First we will find the time required for the ball to reach the ground. Since the ball is thrown horizontally, it has no initial vertical component. We use the following equation to solve for the total flight time:

$\Delta y=v_{iy}t+\frac{1}{2}at^{2}$

We are given the change in height, initial velocity, and acceleration. Using these values, we can solve for the time. Note that the change in height will be negative, since the ball is traveling downward.

$-50m= (0\frac{m}{s})t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-50m=(-4.9\frac{m}{s^2})t^2$

$t=\sqrt{\frac{-50m}{-4.9\frac{m}{s^2}}}$

Finally, we use the horizontal velocity to find the distance traveled in . Remember that the horizontal velocity remains constant during projectile motion.

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

$d=vt=(15\frac{m}{s})(3.19s)=47.9m$

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Question

A is dropped from a height of . A picture is taken when the ball is from the ground with an exposure time of . If the actual diameter of the ball is , what will the vertical diameter of the ball appear to be in the picture?

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Answer

The first step to solving this problem will be to find the velocity of the ball at the point when the picture is taken. We know the initial velocity of the ball (zero), the displacement, and the acceleration. Using the appropriate kinematics formula, we can solve for the final velocity.

$v_f^2=v_o^2+2a\Delta y$

$v_f^2=(0\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(5m-20m)$

$v_f^2=294\frac{m^2}{s^2}$

$v_f=17.15\frac{m}{s}$

Now that we know how fast the ball is traveling when the picture is taken, we can find the distance it travels while the shutter is open. This distance will become a motion blur, making the vertical diameter of the ball appear stretched.

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

$d=(17.15\frac{m}{s})(0.005s)=0.086m$

During the exposure period, the ball will travel , or . The diameter of the ball in the vertical direction will appear to be distorted by this distance.

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Question

A person stands on the edge of a straight -high cliff and holds a ball over the edge. The person tosses the ball directly upward with an initial speed of $5 \frac{m}{s}$ . How long will it take the ball to hit the ground at the base of the cliff, below?

Assume $10 \frac{m}{s^2}$ for gravitational acceleration.

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Answer

This is projectile motion in the vertical direction only, subject to the equation of motion: $(x-x_o) = v_ot + \frac{1}{2}at^2$ .

For this discussion, one can define the downward direction as negative. For projectile motion, $a = -10 \frac{m}{s^2}$ (gravitational acceleration, or ).

In this case, the ball ends up below where is started, so .

The initial velocity, , is $5 \frac{m}{s}$ (upward, thus positive).

With all this, the projectile motion equation becomes:

$- 20 m = (5 \frac{m}{s})t - \frac{1}{2}(10 \frac{m}{s^2})t^2$

This can be solved for using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^{2}-4ac)}) }{2a}$

The result is:

$t = \frac{5\pm 20.62}{10} = 2.562 s$ (or ).

Only the positive answer option is physically possible, and is thus our correct answer.

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Question

A hockey puck of mass is sliding across an ice rink. If the puck loses $2\frac{m}{s}$ of velocity over a distance of , what is the coefficient of kinetic friction between the ice and the puck?

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

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Answer

Since we know the change in velocity of the puck, we can determine the work done by friction by using the work-energy theorem:

$W =Fd= \Delta KE$

The work in this question is done by friction, so we can write:

$F_f d = \Delta KE$

Substituting in expressions for friction and kinetic energy, we get:

$\mu_kF_Nd = \frac{1}{2}m(\Delta v^2)$

The normal force is from gravity, so we can write:

$\mu_kmgd=\frac{1}{2}m(\Delta v^2)$

Rearranging for the coefficient of kinetic friction we get:

$\mu_k = \frac{(\Delta v^2)}{2gd} = \frac{(2\frac{m}{s})^2}{2(10\frac{m}{s^2})(20m)} = 0.01$

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Question

A ball is dropped from a height of . How much time will pass before the ball hits the ground?

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Answer

The equation needed is $x = x_{0}+v_{0}t+\frac{1}{2}at^2$

$0=16m+\frac{1}{2}\left(-10\frac{m}{s^2}\right)t^2$

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Question

Mark, Jim, and David apply forces to an object of mass . The object is on level ground where the coefficient of friction is . Mark applies a force to the right, Jim applies a force to the left, and David applies a force down. How far will the object travel (in meters) after seconds? Assume it starts at rest and the coefficient of static friction is low enough for the object to start moving with the mentioned forces.

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Answer

We can split this question up into the x-direction and y-direction. In the y-direction, the object's acceleration is zero since it's going to remain in contact with the ground. We now can write our $\sum F=ma$ equation (let the up direction be positive). Since it's not accelerating, we have

$\sum F=ma=0\rightarrow F_{n}-mg-3=0$ .

$F_{n}$ denotes the normal force. We know that the mass is and that $g\approx10\frac{m}{s^2}$ , so we can solve for the normal force, which we'll later use to find the force of friction. $F_{n}-(3kg)(10\frac{m}{s^2})-3=0$

$F_{n}=33N$

The force of friction $(F_{f})$ is equal to $\mu F_{n}$ (where $\mu$ denotes the coefficient of friction), so

$F_{f}=(.3)(33N)=9.9N$

In the x-direction, we'll denote right as positive. The forces in the x-direction are right, left, and . Now we can write our $\sum F=ma$ equation as

$\frac{27.1N}{3kg}=9.0\frac{m}{s^2}$

Using this acceleration, we can use kinematics to solve for the distance the object will travel in seconds. We have time and acceleration, but we need distance, so we'll use the equation: $x(t)=(v_{initial})t+0.5at^2$

$v_{initial}=0$

$x=0.5(9.0\frac{m}{s^2})(4s)^2=72m$

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Question

Suppose that an object is dropped from an initial height of 100m above the ground. Neglecting air resistance, how long will it take for this object to reach the ground?

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

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Answer

To solve this problem, we'll need to use a formula that can relate distance to acceleration and time. It's also worth noting that in this case, we are only considering vertical motion along the y-axis. We don't have to worry about horizontal motion along the x-axis.

$\Delta y=v_{i}t+\frac{1}{2}at^{2}$

$\Delta y$ is the vertical displacement

$v_{i}$ is the initial velocity in the vertical direction

is time

is the acceleration due to gravity (which is only in the vertical direction)

Since the object is starting from rest, the initial velocity will be equal to zero, which cancels out the $v_{i}t$ term and gives us:

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

Rearranging to solve for , we obtain:

$t=\sqrt{\frac{2\Delta y}{a}}$

$t=\sqrt{\frac{2(100 m)}{9.8 \frac{m}{s^{2}}}}=4.52s$

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Question

According to the graph shown above, during which interval is Boomer at rest?

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Answer

Boomer is at rest during periods when the slope of the position versus time graph is zero.

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Question

Boomer 's position is shown in the table below.

According to the graph above, during which time interval is Boomer at rest?

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Answer

Boomer's position does not change during the interval when he is at rest. This is the final interval.

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Question

According to the graph above, during which interval does Boomer have the largest speed?

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Answer

Speed is proportional to the absolute value of the slope. Boomer's speed has its largest value $\left(2\frac{m}{s} \right )$ during the intervals from $0s\rightarrow 2.5s$ and from $5s\rightarrow7.5s$ . Note that speed is independent of direction.

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Question

According to the table above, during which interval does Boomer reach his highest speed?

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Answer

Boomer's speed is the ratio of the distance traveled to the time. It is $2\frac{m}{s}$ during both the first and and third intervals.

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Question

According to the graph above, Boomer has the slowest speed during which interval?

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Answer

The smallest speed is zero which occurs during the final intervals, which have a slope of zero.

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Question

According to the graph above, when does Boomer have the smallest average speed?

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Answer

The smallest speed is zero, which occurs during the final intervals which shows no change in position.

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Question

According to the graph above, Boomer has the largest positive velocity during which interval?

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Answer

The largest and only positive velocity occurs during the first interval when the slope is positive.

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