1D Kinematics Problem Solving

The equations of 1D Kinematics are very useful in many situations. While they may seem minimal and straightforward at first glance, a surprising amount of subtlety belies these equations. And the number of physical scenarios to which they can be applied is vast. These problems may not be groundbreaking advances in modern physics, but they do represent very tangible everyday experiences: cars on roads, balls thrown in the air, hockey pucks on ice, and countless more examples can be modeled with these three relatively simple equations.

The three fundamental equations of kinematics in one dimension are:

\[v = v_0 + at,\]

\[x = x_0 + v_0 t + \frac12 at^2,\]

\[v^2 = v_0^2 + 2a(x-x_0).\]

The first gives the change in velocity under a constant acceleration given a change in time, the second gives the change in position under a constant acceleration given a change in time, and the third gives the change in velocity under a constant acceleration given a change in distance.

Here, the subscript "0" always refers to "initial". So, \(v_0\) is the initial velocity, and \(x_0\) is the initial position. Letters with no subscript indicate the quantity value after some time, \(t\). So, in the first equation, \(v\) is the velocity of an object that began at velocity \(v_0\) and has moved with constant acceleration \(a\) for an amount of time \(t\).

Very often, rather than using the initial and final positions, we simply want to know the total change in position, the distance traveled. This change in position is always merely the initial position subtracted from the final position: \(x-x_0\), often called \(d\) for distance. In many problems, this simplifies things and makes it simpler to see what is being asked. With this change, the second and third equations are sometimes rewritten:

\[d = v_0 t + \frac12 at^2,\]

\[v^2 = v_0^2 + 2ad.\]

A ball is dropped from rest off a cliff of height \(100 \text{ m}\). Assuming gravity accelerates masses uniformly on Earth's surface at \(g = 9.8 \text{ m}/\text{s}^2\), how fast is the ball going when it hits the ground? How long does it take to hit the ground?

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Solution:

The third kinematics equation gives the final speed as:

\[v_f^2 = 2( 9.8 \text{ m}/\text{s}^2)(100 \text{ m}) \implies v_f \approx 44.3 \text{ m}/\text{s}.\]

The first kinematical equation gives the time to accelerate up to this speed:

\[t = \frac{v}{a} = \frac{44.3 \text{ m}/\text{s}}{9.8 \text{ m}/\text{s}^2} \approx 4.5 \text{ s}.\]

A soccer ball is kicked from rest at the penalty spot into the net \(11 \text{ m}\) away. It takes \(0.4 \text{ s}\) for the ball to hit the net. If the soccer ball does not accelerate after being kicked, how fast was it traveling immediately after being kicked?

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Solution:

This is a straightforward application of the second equation of motion with \(a = 0\), i.e \(d = vt\):

\[v = \frac{d}{t} = \frac{11 \text{ m}}{0.4 \text{ s}} = 27.5 \text{ m}/\text{s}.\]

A continuously accelerating car starts from rest as it zooms over a span of \(100 \text{ m}\). If the final velocity of the car is \(30 \text{ m}/\text{s}\), what is the acceleration of the car?

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Solution:

Applying the third kinematical equation with \(v_0 = 0\),

\[v^2 = 2ad \implies a = \frac{v^2}{2d} = \frac{900}{200} \text{ m}/\text{s}^2 = 4.5 \text{ m}/\text{s}^2.\]

Sometimes kinematics problems require multiple steps of computation, which can make them more difficult. Below, some more challenging problems are explored.

A projectile is launched with speed \(v_0\) at an angle \(\theta\) to the horizontal and follows a trajectory under the influence of gravity. Find the range of the projectile.

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Solution:

The projectile begins with velocity in the vertical direction of \(v_0 \sin \theta\). To reach the apex of its trajectory, where the projectile is at rest, thus requires a time:

\[t = \frac{v_0 \sin \theta}{g}.\]

The time that it takes to fall back to the ground is therefore double this time,

\[t = \frac{2v_0 \sin \theta}{g}.\]

The range is the total distance in the horizontal direction traveled during this time. This is just the velocity in the x-direction times the time:

\[R = v_x t = v_0 \cos \theta t = \frac{2v_0^2 \sin \theta \cos \theta}{g} = \frac{v_0^2 \sin 2 \theta}{g}.\]

A package is dropped from a cargo plane which is traveling at an altitude of \(10000 \text{ m}\) with a horizontal velocity of \(250 \text{ m}/\text{s}\) and no vertical component of the velocity. The package is initially at rest with respect to the plane. On the ground, a man is speeding along parallel to the plane in a \(5 \text{ m}\) wide car traveling \(40 \text{ m}/\text{s}\) trying to catch the package. The car starts a distance \(X \text{ m}\) ahead of the plane. What does \(X\) need to be for the man to succeed in catching the package?

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Solution:

First, compute how long it takes for the package to hit the ground:

\[d = \frac12gt^2 \implies t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{20000 \text{ m}}{9.8 \text{ m}/\text{s}^2}} = 45.2 \text{ s}.\]

How far does the package travel horizontally during that time?

\[d_{\text{package}} = v_x t = (250 \text{ m}/\text{s})(45.2 \text{ s}) = 11300 \text{ m}.\]

How far does the car travel during that time?

\[d_{\text{car}} = v_x t = (40 \text{ m}/\text{s})(45.2 \text{ s}) = 1808 \text{ m}.\]

If the package is caught, then \(d_{\text{car}} + X = d_{\text{package}}\). This requires:

\[X = (11300 - 1808) \text{ m} = 9492 \text{ m},\]

or nearly \(10\) kilometers!

To be exact, the above quantity for \(X\) can be shifted by up to \(2.5 \text{ m}\) and still make contact with the car, because of the nonzero width of the car, but this is a negligible correction; \(X\) is very large in comparison.