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 is not the same at equal time intervals. Thus, it is dependent on both velocity and time.
You can also learn about Constant Acceleration.
Examples of variable acceleration
Imagine a policeman is chasing a criminal and stumbles on a crowd. The policeman would have to reduce his pace and eventually increase it again once he gets to a less occupied place. The same occurs when a car is accelerating on the highway and meets traffic. The car would accelerate very little in traffic, and when the road becomes free its acceleration increases.
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.
A circular toolbox for variable acceleration problems, Njoku - StudySmart Originals
The function of time in variable acceleration
Remember that variable acceleration is dependent on time because a change in acceleration occurs within time intervals. However, we shall begin by solving examples that expresses velocity and displacement as functions of time.
The displacement s of a particle in motion on a straight line from a point O at time t seconds is given as: \(s = 4t^3 - 9t\) where 0 <t. Find:
a. The displacement when t = 3
b. The time it takes the particle to return to point O.
Solution:
a. In order to find the displacement s, just substitute the value of t = 3 in the equation
\(s = 4t^3 - 9t\)
\(s = 4 (3^3) - 9(3) = 108 - 27 = 81 \space m\)
b. To find the time taken to return to point O, it means the displacement is zero, and would be put in the equation.
Thus:
\(0 = 4t^3 - 9t\)
Factorise
\(0 = t(4t^2 - 9)\)
Let us factories \(4t^2 - 9\) separately
\(4t^2 - 9 = (2t -3)(2t+3)\)
Thus:
\(0 = t (2t - 3) (2t + 3)\)
\(t = 0, 2t - 3 = 0\) or \(2t + 3 = 0\)
\(t = 0, \frac{3}{2} \text{ or } \frac{-3}{2}\)
Recall that 0 <t
That means the answer is \(\frac{3}{2}\) seconds.
The motion of a toy car on a straight track has been modelled to follow the equation \(s = -t^3 + 4t^2\). If this toy leaves at time t = 0 and returns at the start of the track, prove the restriction 0 ≤ t ≤ 4.
Solution:
From the start of the motion,
s ≥ 0
So,
\(-t^3 + 4t^2 \geq 0, \quad 4t^2 - t^3 \geq 0, \quad t^2(4-t) \geq 0, \qquad 4-t \geq 0\)
\(t^2 \geq 0\) and \(4 \geq t\)
Since time cannot be in negative. We can consider only t ≥ 0 in this case.
thus; t ≤ 4
t ≥ 0 and t ≤ 4
which proves the restriction 0 ≤ t ≤ 4.
The relationship between the velocity and time of an object moving in a straight line is given by the expression: \(v = 2t^2 - 16t + 24\) for t ≥ 0
Find:
- The initial velocity.
- The time the object is instantaneously at rest.
- The time the velocity is at 64 m/s.
- The greatest speed with the interval 0 ≤ t ≤ 5.
Solution
a. At the initial velocity the time is zero because the motion just began.
t = 0
substitute the value of t in the equation to find the velocity
\(v = 2t^2 - 16t + 24\)
v = 24 m / s.
b. At the instantaneous velocity at rest, v = 0
Substitute that in the equation to find the time.
\(0 = 2t^2 - 16t + 24 = 2(t^2-8t+12)\)
Divide both sides by 2
\(0 = t^2 - 8t + 12\)
Factorise
\(0 = (t - 6) (t - 2)\)
t = 6 or 2
Thus, the object comes to an instantaneous rest at 2 seconds and 6 seconds.
c. At the velocity = 64 m / s, the time is:
\(64 = 2t^2 - 16t+24 = 2 (t^2 - 8t + 12)\)
Divide both sides by 2
\(32 = t^2 - 8t + 12\)
\(0 = t^2-8t + 12 -32 = t^2 - 8t - 20\)
Factorise
\(0 = (t - 10) (t + 2)\)
t = 10 or -2
Thus, the object moves at 64 m/s after 10 seconds.
d. The greatest speed between 0 ≤ t ≤ 5 is the highest speed attained by the object within t = 0 and t = 5.
To derive this, a graph of the equation could be plotted with the values between 0 to 5, with v in the y-axis and t in the x-axis.
Initially, the value for t = 0 has been derived in question a) as 24 m / s. Use the same approach and fill in the table for values 1 to 5.
\(x = t\) | 0 | 1 | 2 | 3 | 4 | 5 |
\(y = v = 2t^2 - 16t+24\) | 24 | 10 | 0 | -6 | -8 | -6 |
The graph below reveals the max velocity. This is seen in the table and confirmed by the graph that the highest velocity is 24 m / s. This is obtained at time t = 0.
A velocity-time graph showing the max velocity, Njoku - StudySmarter Originals
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. Let's look at some examples to make this clearer:
A tricycle is moving horizontally. The displacement q from rest O at time t is given as:
\(q = t^4-32t+12\)
a. Calculate the velocity of the tricycle when t is 4 seconds.
b. Find the time it attains an instantaneous velocity.
c. Calculate the acceleration when t is 1.5 seconds.
Solution:
a. q is the displacement given as
\(q = t^4-32t+12\)
Remember that you differentiate the displacement to get the velocity.
\(v = \frac{dq}{dt} = \frac{d(q = t^4 -32t+12)}{dt}\)
\(v = 4t^3 -32\)
Since the t function of the velocity has been derived, substitute the value of t = 4 in the equation.
\(v = 4(4^3)-32 = 256-32 = 224\)
Therefore the velocity of the tricycle is 224 m/s
b. To find the time at which the tricycle experiences instantaneous velocity, always remember that the object is momentarily at rest. It means that velocity is zero at that instant.
Input the value of v = 0 in the equation \(v = 4t^3 - 32\) to find time t.
\(0 = 4t^3 - 32\)
\(4t^3 = 32\)
Divide both sides by 4
\(t^3= 8\)
Find the cube root of both sides: t = 2
Therefore the tricycle experiences instantaneous velocity after 2 s.
c. Remember to differentiate your velocity when determining your acceleration.
Since \(v = 4t^3 - 32\)
\(a = \frac{dv}{dt} = \frac{d(v=4t^3-32)}{dt}\)
\(a = 12t^2\)
Now you are asked to find the acceleration. Just substitute the value of t = 1.5 in the acceleration equation.
\(a = 12(1.5^2); \quad a = 27 ms^{-2}\)
Therefore the acceleration at time t = 1.5 s is 27 ms-2
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.
A woman has a spring that leaves her hand at time t = 0 seconds, and moves vertically in a straight line before returning back to her hand. The distance y between the spring and her hand after t seconds is given by:
\(y = -0.2t^3 + 0.4t^2 + 0.6 t\), 0 ≤ t ≤ 3
a. Prove the restriction 0 ≤ t ≤ 3.
b. Calculate the maximum distance between the woman's hand and the spring.
Solution:
a. \(y = -0.2t^3 + 0.4t^2 + 0.6 t\)
Factorise:
\(y = -0.2t(t^2 - 2t - 3)\)
Factorise further:
\(y = -0.2t (t - 3) (t + 1)\)
when y = 0, then
\(-0.2t = 0, \space t - 3 = 0 \text{ or } t + 1 = 0\)
t = 0, 3 or -1
There are no negative values for time, so, t = 0 or 3. This justifies the restriction of t, 0 ≤ t ≤ 3.
b. Remember that when a particle reaches its maximum displacement, it becomes momentarily at rest. We say its velocity is instantaneously at rest and v = 0.
Thus, to find the max displacement, we need to know the time the object becomes instantaneously at rest. To do this, we need to find the t function of v.
\(y = -0.2t^3+0.4t^2+0.6t\)
\(v = \frac{dy}{dt} = \frac{d(y = -0.2t^3 + 0.4 t^2 + 0.6t)}{dt}\)
\(v = -0.6t^2 + 0.8t + 0.6\)
Now we have the t function of our velocity in the equation, let us find t when v is instantaneously at rest.
v = 0
\(0 = -0.6t^2 + 0.8t + 0.6\)
Factorise
\(0 = -0.2(3t^2-4t-3)\)
Divide both sides by -0.2
\(0 = 3t^2-4t-3\)
Using the formula for quadratic equations \(\frac{-b \pm \sqrt{b^2 -4ac}}{2a}\)
Where a = 3, b = -4 and c = -3
After substituting the values and solving the equation,
t = 1.8685 or -0.5351.
Remember no negative value of t is valid, thus, t = 1.8685
This means the distance y is maximum after 1.8685 seconds.
Substitute the value of t to find ymax.
\(y = -0.2t^3 + 0.4t^2 + 0.6t\)
\(y_{max} = -0.2(1.8685)^3 + 0.4(1.8685)^2 + 0.6(1.8685) = 1.2129\)
The maximum distance between the woman's hand and the spring is 1.21 meters.
Using integration in variable acceleration problems
Initially, we have seen how differentiation 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, integration is used.
You know that:
\(v = \frac{ds}{dt}\) other \(a = \frac{dv}{dt}\)
Then:
\(s = \int{v \space dt}\) other \(v = \int{a \space dt}\)
\((t-5)ms^{-2}\) is the acceleration of a basketball bounced by a teenager. The ball starts with a velocity of 8 m/s. Determine the time (s) the basketball is instantaneously at rest.
Solution:
An illustration of the basketball instantaneously at rest, Njoku - StudySmarter Originals
\(a = t - 5\)
\(v = \int{a \space dt}\)
\(v = \int{(t-5) dt} = \frac{t^2}{2} - 5t +c\) do not forget to add the constant c after each integration.
We were told that the ball began with a velocity of 8 m / s, which means that at time t = 0, v = 8
Thus substitute the value of t and v in the equation \(8 = \frac{0^2}{2} - 5(0) + c\)
c = 8
This means the t function of the velocity is:
\(v = \frac{t^2}{2} - 5t + 8\)
Now we have a complete equation for the velocity, we can determine the time (s) the ball is instantaneously at rest.
v = 0 because the object is momentarily at rest.
\(v = \frac{t^2}{2} - 5t + 8\)
\(0 = \frac{t^2}{2} - 5t +8\)
Multiply the equation by 2 to get rid of the fraction
\(0 = t^2 - 10t + 16\)
Factorise
\((t - 8) (t - 2) = 0\)
t = 8 or 2
Thus, the basketball is instantaneously at rest in 2 and 8 seconds.
Variable Acceleration - Key takeaways
- Variable acceleration takes place when velocity changes are not equal for the same time intervals. Thus, the acceleration is not constant.
- Differential and Integral Calculus are tools used to solve problems in variable acceleration.
- You differentiate to find velocity when the displacement has been expressed as a function of time. The same applies to acceleration when velocity is given.
- You integrate to find displacement when the velocity has been expressed as a function of time. The same applies to velocity when acceleration is given.
- The velocity of an object instantaneously at rest is 0.
- You can find the maximum displacement by differentiating the displacement to find the t function of the velocity. Then take the time at which the velocity is instantaneously at rest and substitute it in the displacement equation.
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