If an object moves at a non-zero constant acceleration for a certain interval of time, then the distance it covers in that time

Question

Prepp Team · Accepted Answer

Understanding Distance with Constant Acceleration

Let's analyze how the distance an object covers is determined when it moves with a non-zero constant acceleration over a specific time interval. This is a fundamental concept in kinematics, the study of motion.

When an object moves with constant acceleration, its motion can be described using kinematic equations. One of the key equations that relates displacement, initial velocity, acceleration, and time is:

\begin{equation*} s = ut + \frac{1}{2}at^2 \end{equation*}

Where:

\(s\) represents the displacement (change in position) of the object during the time interval. In the case of motion in a single direction, this is also the distance covered.
\(u\) represents the initial velocity of the object at the start of the time interval.
\(a\) represents the constant acceleration of the object.
\(t\) represents the duration of the time interval.

The question asks about the distance the object covers in that time interval. Looking at the equation \(s = ut + \frac{1}{2}at^2\), we can see how the displacement (and thus distance covered if motion is in one direction) depends on the variables on the right side of the equation: \(u\), \(a\), and \(t\).

Analyzing the Dependence of Distance Covered

The term \(ut\) clearly shows that the displacement \(s\) is directly influenced by the initial velocity \(u\). If the initial velocity changes, the displacement (and distance covered) will also change, assuming acceleration and time remain constant.

Let's consider the given options based on this understanding:

Option 1: depends on its initial velocity - The equation \(s = ut + \frac{1}{2}at^2\) includes the term \(ut\). This term directly links the displacement (and distance covered in many cases) to the initial velocity \(u\). Therefore, the distance covered depends on the initial velocity. This statement is consistent with the kinematic equation.
Option 2: is independent of its initial velocity - This contradicts the presence of the \(ut\) term in the kinematic equation. The initial velocity clearly affects the distance covered.
Option 3: increases linearly with time - The equation includes a term with \(t^2\) (\(\frac{1}{2}at^2\)) in addition to the term with \(t\) (\(ut\)). This indicates that the relationship between displacement and time is generally quadratic, not purely linear, unless the acceleration \(a\) is zero. Since the question states non-zero constant acceleration, the dependence on time is not strictly linear.
Option 4: depends on its initial displacement - The equation \(s = ut + \frac{1}{2}at^2\) calculates the *displacement* (\(\Delta x\)), which is the *change* in position. It does not require knowing the initial position (\(x_i\)) itself, only the initial velocity, acceleration, and time interval. The initial displacement (position) determines the final position (\(x_f = x_i + s\)) but not the displacement or distance covered during the interval.

Based on the kinematic equation for motion with constant acceleration, the distance covered over a certain time interval is dependent on the initial velocity, the constant acceleration, and the duration of the time interval.

Conclusion on Distance and Initial Velocity

Therefore, if an object moves at a non-zero constant acceleration for a certain interval of time, the distance it covers in that time depends on its initial velocity.

Variable	Role in Kinematic Equation (\(s = ut + \frac{1}{2}at^2\))	Does Distance Covered Depend On It?
Initial Velocity (\(u\))	Multiplier of the time term (\(t\))	Yes (Explicitly in \(ut\))
Constant Acceleration (\(a\))	Multiplier of the time-squared term (\(t^2\))	Yes (Explicitly in \(\frac{1}{2}at^2\))
Time Interval (\(t\))	Present in both terms (\(t\) and \(t^2\))	Yes (Explicitly in \(ut\) and \(\frac{1}{2}at^2\))
Initial Displacement (\(x_i\))	Not present in the equation for displacement \(s\)	No (Displacement is change in position)

Revision Table: Constant Acceleration Concepts

Concept	Description	Key Equation
Displacement (\(s\) or \(\Delta x\))	Change in position vector. In 1D, final minus initial position.	\(s = x_f - x_i\)
Velocity (\(v\))	Rate of change of position (speed with direction).	\(v = \frac{ds}{dt}\)
Acceleration (\(a\))	Rate of change of velocity. Constant means \(a\) is constant over time.	\(a = \frac{dv}{dt}\)
Kinematic Equation 1	Velocity as a function of initial velocity, acceleration, and time.	\(v = u + at\)
Kinematic Equation 2	Displacement as a function of initial velocity, acceleration, and time.	\(s = ut + \frac{1}{2}at^2\)
Kinematic Equation 3	Velocity squared as a function of initial velocity, acceleration, and displacement.	\(v^2 = u^2 + 2as\)

Additional Information: Distance vs. Displacement

It's important to distinguish between distance covered and displacement. Displacement (\(s\)) is the straight-line distance from the initial point to the final point, with direction. Distance covered is the total length of the path traveled.

In motion with constant acceleration in one dimension, if the object does not change direction during the time interval, the distance covered is equal to the magnitude of the displacement, \(|s|\). The formula \(s = ut + \frac{1}{2}at^2\) gives the displacement.

If the object changes direction during the interval (which happens if the velocity becomes zero at some point within the interval, and initial velocity and acceleration have opposite signs), then calculating the total distance covered requires finding the time when velocity is zero (\(t_{turn}\)) using \(v = u + at = 0\). The total distance is then the sum of the magnitudes of displacements in the two parts of the motion (from \(0\) to \(t_{turn}\) and from \(t_{turn}\) to the end of the interval \(t\)). Even in this more complex case, the initial velocity (\(u\)) is crucial in determining if a turn happens and where, thus directly affecting the total distance covered.

The question asks what the distance covered depends on. Since the displacement \(s\) depends on initial velocity, and distance covered is directly related to displacement (either equal magnitude or sum of magnitudes of segments of displacement, both of which are affected by \(u\)), the distance covered depends on the initial velocity.