# 1D Kinematics Problem Solving

Below, the equations of one-dimensional kinematics are briefly reviewed and a number of practice problems are presented.

## Equation Review

The three fundamental equations of kinematics in one dimension are:

\[v = v_0 + at,\]

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

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

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.

## 1D Kinematics Problems: Easy

## 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?

Solution:

The third kinematical 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?

Solution:

This is a straightforward application of the first 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?

Solution:

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

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

## 1D Kinematics Problems: Medium

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.

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?

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.

A pitcher throws a baseball towards home plate, a distance of \(18 \text{ m}\) away, at \(v = 40 \text{ m}/\text{s}\). Suppose the batter takes \(.2 \text {s}\) to react before swinging. In swinging, the batter accelerates the end of the bat from rest through \(2 \text{ m}\) at some constant acceleration \(a\). Assuming that the end of the bat hits the ball if it crosses the plate within \(. 05 \text{ s}\) of the ball crossing the plate, what is the minimum required \(a\) in \(\text{m}/\text{s}^2\) to the nearest tenth for the batter to hit the ball?

SpaceX is trying to land their next reusable rocket back on a drone ship. The drone ship is traveling due west in an ocean current at a constant speed of \(5 \text{ m}/\text{s}\). The rocket is \(2000 \text{ m}\) east of the drone ship and \(5000 \text{ m}\) vertically above it, traveling vertically downwards at \(100 \text{ m}/\text{s}\). If the rocket can apply vertical and horizontal thrusts to change the acceleration of the rocket in the vertical and horizontal directions, and must accelerate constantly in both directions, find the **magnitude** of the net acceleration (vector) required to land on the ship with no vertical velocity. Answer in \(\text{ m}/\text{s}^2\) to the nearest tenth.

**Cite as:**1D Kinematics Problem Solving.

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