Solving rectilinear motion problems
What is rectilinear motion, and which formulas do we use to model it?
Rectilinear motion problems deal with an object that moves laterally, or horizontally.
The object can be moving along the ground or at any other height, as long as it’s moving horizontally. We call this type of motion “rectilinear” motion.
Problems like these require you to know the relationship between position ???x(t)???, velocity ???v(t)???, and acceleration ???a(t)???. The important thing to know is that the derivative of position is velocity, and the derivative of velocity is acceleration.
???x(t)???
???x'(t)=v(t)???
???x''(t)=v'(t)=a(t)???
We can also describe the above relationship using integrals instead of derivatives, and we see that the integral of acceleration is velocity, and the integral of velocity is position.
???a(t)???
???\int{a(t)}\ dt=v(t)???
???\int\int{a(t)}\ dt=\int v(t)=x(t)???
How to solve rectilinear motion problems
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How to model the position of an object that’s moving along the ground
Example
An object is moving along the ground. Its acceleration is ???a(t)=3t+5???, its velocity at time ???t=4??? is ???v(4)=6???, and its position at ???t=5??? is ???x(5)=25???. Find the equation for position that describes this object’s motion.
We can integrate the acceleration function to get a velocity function, ???v(t)???.
???v(t)=\int a(t)\ dt???
???v(t)=\int3t+5\ dt???
???v(t)=\int3t\ dt+\int5\ dt???
???v(t)=3\int{t}\ dt+5\int1\ dt???
???v(t)=\frac{3t^2}{2}+5t+C???
Now we need to solve for ???C??? using ???v(4)=6???.
???6=\frac{3(4)^2}{2}+5(4)+C???
???C=-38???
So the equation for velocity, ???v(t)???, is
???v(t)=\frac{3t^2}{2}+5t-38???
Now we can integrate the velocity function to get the position function.
???x(t)=\int{v(t)}\ dt???
???x(t)=\int\frac{3t^2}{2}+5t-38\ dt???
???x(t)=\int\frac{3t^2}{2}\ dt+\int5t\ dt+\int-38\ dt???
???x(t)=\frac{3}{2}\int{t^2}\ dt+5\int{t}\ dt-38\int 1\ dt???
???x(t)=\frac{3}{2}\left(\frac{t^3}{3}\right)+5\left(\frac{t^2}{2}\right)-38t+C???
???x(t)=\frac{t^3}{2}+\frac{5t^2}{2}-38t+C???
Now we need to solve for ???C??? using ???x(5)=25???.
???25=\frac{(5)^3}{2}+\frac{5(5)^2}{2}-38(5)+C???
???C=90???
So the equation for position, ???x(t)???, is
???x(t)=\frac{t^3}{2}+\frac{5t^2}{2}-38t+90???