We previously explored kinematics and used differentiation to solve problems involving the motion of objects in a straight line. Let's now extend our skills to look at applications in kinematics requiring integration.
The displacement function, $x\left(t\right)$x(t), gives the position of a particle from the origin at time $t$t.
The velocity function, $v\left(t\right)$v(t), gives the instantaneous rate of change of the displacement function with respect to time.
The acceleration function, $a\left(t\right)$a(t), gives the instantaneous rate of change of the velocity with respect to time.
We have the following relationships:
$v\left(t\right)=x'\left(t\right)$v(t)=x′(t)
and
$a\left(t\right)=v'\left(t\right)=x''\left(t\right)$a(t)=v′(t)=x′′(t)
Given the displacement function, we can obtain the velocity and in turn the acceleration function through differentiation. Given the velocity function, we can use integration to find the net change in displacement and given additional information, such as the initial position, we can find the displacement function. Similarly, given the acceleration function, we can use integration to find the net change in velocity and given additional information, we can find the velocity function. These relationships are illustrated in the following diagram:
To find net displacement, we can integrate the velocity function over a given interval. Net displacement may include both positive and negative values, as the velocity function accounts for both forward and backward movement. Total distance travelled, on the other hand, is always positive. To find the total distance travelled by an object, regardless of direction, we need to integrate the absolute value of the velocity function.
Given the velocity function, $v\left(t\right)$v(t), if we wish to calculate the net change in displacement between $t=a$t=a and $t=b$t=b we have:
To calculate the distance travelled, that is the net change in distance between $t=a$t=aand $t=b$t=b we have:
Note: To complete this calculation we could use the calculator or alternatively, we can find the area under our velocity curve, thus taking into account any negative portions.
Consider a body moving so that its velocity, in m/s, is a function of time is given by $v(t)=6t^2-6t-12$v(t)=6t2−6t−12. The object's initial displacement is $-4$−4 m.
(a) Find its displacement function $x(t)$x(t).
Think: Using integration and the additional information that $x(0)=-4$x(0)=−4 we can find the displacement function.
Do:
$x(t)$x(t) | $=$= | $\int6t^2-6t-12\ dt$∫6t2−6t−12 dt |
$=$= | $2t^3-3t^2-12t+c$2t3−3t2−12t+c |
We know that $x(0)=-4$x(0)=−4. Therefore, $c=-4$c=−4 and the displacement function is given by $x(t)=t^3-3t^2-12t-4$x(t)=t3−3t2−12t−4.
(b) What is the displacement of the object at $t=3$t=3?
Think: Substitute $t=3$t=3 into the displacement function.
Do:
$x(3)$x(3) | $=$= | $2(3)^3-3(3)^2-12(3)-4$2(3)3−3(3)2−12(3)−4 |
$=$= | $-13$−13 |
The object is $13$13 m to the left of the origin.
(c) Find the net displacement of the object in the first $5$5 seconds?
Think: We could evaluate our displacement function found in part (a) at $0$0 and $5$5 and find the difference or evaluate the definite integral of the velocity function over the interval.
Do:
Net change in displacement | $=$= | $\int_0^5\left(6t^2-6t-12\right)dt$∫50(6t2−6t−12)dt |
$=$= | $\left[2t^3-3t^2-12t\right]_0^5$[2t3−3t2−12t]50 | |
$=$= | $115-0$115−0 | |
$=$= | $115$115 |
Thus, the net displacement in the first $5$5 seconds is $115$115 m.
(d) Find the distance travelled by the object in the first $5$5 seconds?
Think: To calculate the distance travelled we need to integrate the absolute value of the velocity function over the given interval.
Do: The integral $\int_0^5|v(t)|\ dt$∫50|v(t)| dt can be calculated using our calculator or by finding the area under the velocity function. We can see in the graph below, or find algebraically, that the velocity function crosses the horizontal axis at $t=2$t=2. Hence, we can find the area by splitting the integral into the two regions $A_1$A1 and $A_2$A2.
Net change in distance | $=$= | $A_1+A_2$A1+A2 |
$=$= | $-\int_0^2\left(6t^2-6t-12\right)dt+\int_2^5\left(6t^2-6t-12\right)dt$−∫20(6t2−6t−12)dt+∫52(6t2−6t−12)dt | |
$=$= | $-\left[2t^3-3t^2-12t\right]_0^2+\left[2t^3-3t^2-12t\right]_2^5$−[2t3−3t2−12t]20+[2t3−3t2−12t]52 | |
$=$= | $-\left(-20-0\right)+115+20$−(−20−0)+115+20 | |
$=$= | $155$155 |
Thus, the distance travelled in the first $5$5 seconds in $155$155 m.
Reflect: Under what circumstances would the net change in displacement and distance travelled be equal?
A falling body near the surface of the earth has a constant acceleration of about $-9.8$−9.8 m/s^{2}. If the body starts from rest at $t=0$t=0 and begins its fall from a height of $200$200 m, find its displacement function.
Think: Using integration and the additional information that $v(0)=0$v(0)=0 we can find the velocity function. We can then use integration once more and the initial position to find the displacement function.
Do: Acceleration is the rate of change of velocity. So using $v(t)=\int a(t)\ dt$v(t)=∫a(t) dt, we have:
$v(t)$v(t) | $=$= | $\int(-9.8)\ dt$∫(−9.8) dt |
$=$= | $-9.8t+c$−9.8t+c |
We were given that the body started from rest, that is it had an initial velocity of $0$0 m/s. So we have:
$v(0)$v(0) | $=$= | $-9.8\times0+c$−9.8×0+c |
$0$0 | $=$= | $c$c |
Hence, the velocity function is $v(t)=-9.8t$v(t)=−9.8t.
Using that displacement is given by $x(t)=\int v(t)\ dt$x(t)=∫v(t) dt, we obtain:
$x(t)$x(t) | $=$= | $\int-9.8t\ dt$∫−9.8t dt |
$=$= | $\frac{-9.8t^2}{2}+c$−9.8t22+c | |
$=$= | $-4.9t^2+c$−4.9t2+c |
Using the information at $t=0$t=0, $x=200$x=200. We can see $c=200$c=200 and the displacement function is $x(t)=-4.9t^2+200$x(t)=−4.9t2+200.
The velocity of a particle moving in a straight line is given by $v\left(t\right)=6t+15$v(t)=6t+15, where $v$v is the velocity in metres per second and $t$t is the time in seconds.
The displacement after $2$2 seconds is $45$45 metres to the right of the origin.
Calculate the initial velocity of the particle.
Determine the function $x\left(t\right)$x(t) for the position of the particle. Use $C$C as the constant of integration.
Calculate the displacement of the particle after four seconds.
Find the total distance travelled between two and four seconds.
Consider the velocity function $v\left(t\right)=5\left(1-e^{-t}\right)$v(t)=5(1−e−t), where $v$v is in m/s and $t$t is in seconds.
Calculate the total change in displacement over the interval $\left[0,5\right]$[0,5]. Leave your answer in terms of $e$e, in factorised form.
Hence, calculate the average change in displacement over the interval $\left[0,5\right]$[0,5]. Give your answer correct to three decimal places where appropriate.
Use your CAS calculator to calculate the average change in the distance over the interval $\left[0,5\right]$[0,5]. Give your answer correct to three decimal places where appropriate.
Why was the average change in displacement equal to the average change in distance? Choose all appropriate answers.
Because the velocity is always positive.
Because the displacement function never crosses the $x$x-axis
Because the displacement function is always increasing.
Because $1-e^0$1−e0 is equal to 0.
The acceleration $a$a (in m/s^{2}) of an object travelling horizontally along a straight line after $t$t seconds is modelled by $a\left(t\right)=6t-27$a(t)=6t−27, where $t\ge0$t≥0.
After $10$10 seconds, the object is moving at $90$90 metres per second in the positive direction.
State the velocity $v$v of the particle at time $t$t. Use $C$C as the constant of integration.
Solve for all times $t$t at which the particle is at rest. If there are multiple solutions write them on the same line separated by commas.
The acceleration of a particle is given by $a\left(t\right)=2e^{3t}$a(t)=2e3t m/s², and its velocity is $10$10 m/s initially.
Find its velocity $v\left(t\right)$v(t), after $t=2$t=2 seconds. Give your answer correct to the nearest metre per second.
Let $x\left(t\right)$x(t) be the displacement of the particle at $t$t seconds. If its initial position is $2$2 m to the left of the origin, find its displacement after $4$4 seconds, correct to the nearest metre.
Apply differentiation and integration to kinematics problems that involve displacement, velocity and acceleration of a particle moving in a straight line with variable or constant acceleration, and the use of x–t and v–t graphs.