Variable acceleration is a situation where there is a difference in the average acceleration within different points along the path of an object in motion. Differences could be either in magnitude, direction, or in both magnitude and direction. Variable acceleration occurs when changes in the velocity of an object are not the same at equal time intervals. Thus, it is dependent on both velocity and time.
Examples of Variable Acceleration
Consider a cyclist riding uphill and then transitioning to a downhill slope. The cyclist's acceleration decreases while struggling uphill due to increased resistance and then significantly increases when going downhill due to gravity.
Similarly, when a rocket is launched into space, its acceleration changes at different stages of the ascent. Initially, the rocket accelerates rapidly as it burns fuel to escape Earth's gravity, but as it reaches higher altitudes and expends fuel, its acceleration varies due to decreasing gravitational pull and reduced mass.
Tools Needed in Variable Acceleration
In order to solve variable acceleration questions, you must have a background knowledge of calculus. This is essential because in order to derive velocity when displacement has been given, you are expected to differentiate the displacement. Meanwhile, to determine your displacement with a given velocity, you are expected to integrate the velocity. This process is repeated between velocity and acceleration and vice versa.
The Role of Time in Variable Acceleration
Variable acceleration is directly related to time, as changes in acceleration occur over specific time intervals.
When dealing with variable acceleration problems, it is important to consider how velocity and displacement are expressed as functions of time. By understanding these relationships, we can effectively solve examples and analyze the impact of changing acceleration on an object's motion.
Example: Calculating Distance with Variable Acceleration
Consider an object moving along a straight line with a variable acceleration given by \(a(t) = 3t^2 + 2t\), where \(a(t)\) is the acceleration in meters per second squared (\(m/s^2\)) and \(t\) is the time in seconds. If the object starts from rest at \(t = 0\), calculate the distance it covers in the first 2 seconds.
Solution:
To find the distance covered, we first need to determine the velocity function \(v(t)\) by integrating the acceleration function \(a(t)\).
\(v(t) = \int a(t) \, dt = \int (3t^2 + 2t) \, dt = t^3 + t^2 + C\)
Since the object starts from rest, \(v(0) = 0\), which implies \(C = 0\). Therefore, \(v(t) = t^3 + t^2\).
Next, we find the distance covered, \(s(t)\), by integrating the velocity function.
\(s(t) = \int v(t) \, dt = \int (t^3 + t^2) \, dt = \frac{1}{4}t^4 + \frac{1}{3}t^3 + D\)
Since the object starts from a specific point, we can assume \(s(0) = 0\), which implies \(D = 0\). Thus, \(s(t) = \frac{1}{4}t^4 + \frac{1}{3}t^3\).
Finally, to find the distance covered in the first 2 seconds, we calculate \(s(2)\):
\(s(2) = \frac{1}{4}(2)^4 + \frac{1}{3}(2)^3 = 4 + \frac{8}{3} = \frac{20}{3}\)
Therefore, the object covers \(\frac{20}{3}\) meters, or approximately 6.67 meters, in the first 2 seconds.
Applications of Differentiation in Variable Acceleration
Velocity is defined as the change in displacement with time. We have explained the function of time in displacement and velocity, thus, \(v = \frac{ds}{dt}\).
In a similar way, acceleration is defined as the change in velocity with time. If velocity is overexpressed as a function of t to derive the acceleration, then, \(a = \frac{dv}{dt} = \frac{d^2 s}{dt^2}\).
This knowledge of differential calculus would be needed when determining the velocity if the displacement is expressed as a function of time. In the same way, the acceleration can be derived when the velocity is expressed as a function of time.
Example: If the displacement of an object is given by the function \(s(t) = 4t^2 + 2t\), where \(s\) is in meters and \(t\) is in seconds, find the velocity and acceleration of the object at any time \(t\).
Solution:
- To find the velocity (\(v\)), we differentiate the displacement (\(s\)) with respect to time (\(t\)), which gives us: \[v = \frac{ds}{dt} = \frac{d}{dt}(4t^2 + 2t) = 8t + 2.\]
- Then, to find the acceleration (\(a\)), we differentiate the velocity (\(v\)) with respect to time (\(t\)), which yields: \[a = \frac{dv}{dt} = \frac{d}{dt}(8t + 2) = 8.\]
Therefore, at any given time \(t\), the velocity of the object is \(8t + 2\) meters per second, and the acceleration is constant at \(8\) meters per second squared.
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Using Variable Acceleration in Minimum and Maximum Instances
The idea of variable acceleration also has profitable applications in finding the minimum and maximum values of displacement, velocity and acceleration.
Example:
An object moves along a straight line with its acceleration given by \(a(t) = 12 - 6t\), where \(a\) is in meters per second squared and \(t\) is time in seconds. Determine the time when the object reaches its maximum velocity if it starts from rest at \(t = 0\).
Solution:
- Since the object starts from rest, the initial velocity \(v_0 = 0\) m/s at \(t = 0\).
- To find the velocity as a function of time, integrate the acceleration: \[v(t) = \int a(t) \, dt = \int (12 - 6t) \, dt = 12t - 3t^2 + C.\]
- Using the initial condition \(v(0) = 0\), we find \(C = 0\), so \(v(t) = 12t - 3t^2\).
- To find the maximum velocity, we need to determine when the acceleration is zero because the velocity reaches a maximum or minimum when \(a(t) = 0\): \[12 - 6t = 0 \Rightarrow t = 2 \, \text{seconds}.\]
- To confirm that this is a maximum, we can check the second derivative of velocity (which is the derivative of acceleration) or simply observe that the acceleration changes from positive to negative at \(t = 2\), indicating a maximum point.
Therefore, the object reaches its maximum velocity at \(t = 2\) seconds.
Using Integration in Variable Acceleration Problems
Initially, we have seen how is used to find the velocity from a given displacement with respect to time. We have also seen how this process is used to find the acceleration when the velocity is given as a function of time. To carry out the reverse of these situations such as from velocity to displacement as well as acceleration to velocity, is used.
Example:
An object is moving in a straight line with an acceleration that varies with time as \(a(t) = 4t\), where \(a\) is in meters per second squared and \(t\) is time in seconds. If the initial velocity of the object at \(t = 0\) is 2 m/s, find the velocity of the object at any time \(t\) and its displacement at \(t = 2\) seconds.
Solution:
- First, find the velocity as a function of time by integrating the acceleration: \[v(t) = \int a(t) \, dt = \int 4t \, dt = 2t^2 + C.\]
- Given that the initial velocity \(v(0) = 2\) m/s, we can solve for \(C\): \[2 = 2(0)^2 + C \Rightarrow C = 2.\] Therefore, \(v(t) = 2t^2 + 2\).
- To find the displacement at \(t = 2\) seconds, we integrate the velocity function: \[s(t) = \int v(t) \, dt = \int (2t^2 + 2) \, dt = \frac{2}{3}t^3 + 2t + D.\]
- If we assume the initial displacement \(s(0) = 0\) for simplicity (or it's not given), then \(D = 0\), and thus: \[s(t) = \frac{2}{3}t^3 + 2t.\] Plugging in \(t = 2\) seconds: \[s(2) = \frac{2}{3}(2)^3 + 2(2) = \frac{16}{3} + 4 = \frac{28}{3} \, \text{meters}.\]
Therefore, the velocity of the object at any time \(t\) is \(2t^2 + 2\) m/s, and its displacement at \(t = 2\) seconds is \(\frac{28}{3}\) meters.
Variable Acceleration - Key takeaways
Variable acceleration signifies that an object's velocity changes inconsistently across equal time periods, leading to a non-uniform acceleration rate. Understanding and solving variable acceleration scenarios heavily rely on Differential and Integral Calculus. Differentiation is employed to ascertain velocity from a known time-dependent displacement function, and similarly, to derive acceleration from a given velocity function.
Conversely, integration is utilized to calculate displacement from velocity, and to determine velocity from known acceleration. An object is considered instantaneously at rest when its velocity drops to zero. To calculate maximum displacement, one must differentiate the displacement function to obtain velocity as a function of time, pinpoint when the velocity is zero (indicating instantaneous rest), and then apply this time value back into the displacement formula.