# Calculus problems involving motion

I've been working on the following problems, and I know how to integrate functions,but I do not know how to find the value of "c" in the examples below when finding the antiderivative. Any idea what to do? Cheers

1. A particle travels in a straight line such that its acceleration at time t seconds is equal to $6t+1$ $m/s^2$. When $t=2$, the displacement is equals to $12m$ and when $t=3$ the displacement is equals to $34m$. Find the displacement and velocity when $t=4$.

2. A particle travels in a straight line with its acceleration at time t equal to $3t+2$ $m/s^2$. The particle has an initial positive velocity and travels $30m$ in the fourth second. Find the velocity of the body when $t=5$.

• Are you familiar with solving systems of two equations of two unkonws ?? – Nizar Nov 26 '15 at 9:00
• No, not really, however I think that there should be a simpler way of solving this problem... – John Nov 26 '15 at 9:05
• I'm pretty sure that that was covered in the first week of grade 10. – John Joy Nov 26 '15 at 14:46
• Yeah, but im in grade 9 right now.... – John Nov 27 '15 at 0:34

## 3 Answers

For the first problem: you have the acccelaration is $$a(t)=6t+1$$ Let $x(t)$ and $v(t)$ denote the displacement and the velocity respectively. Then $x''(t)=a(t)$. So integration $a(t)$ twice we end up with $$x(t)= t^3 +\frac{t^2}{2}+At+B$$ but we have $x(2)=12$ and $x(3)=34$ Then $$8+ 2 +2A+B= 12$$ and $$27+\frac{9}{2}+3A+B=34$$ subtracting the first equation from the second we get $17+\frac{9}{2} +A= 22$, hence $A= \frac{1}{2}$.Then substitute to find $B$.

HINT: Your first problem requires you to find both displacement and velocity. So you have to solve the first order as well as the second order differential equation. For the second order differential equation, you will have 2 constants, say $A$ and $B$. Use the 2 conditions for displacement and solve for the constants.Then differentiate to get the expression for velocity.

• For #1....A=6t+1 therefore v= 3t² + t+ c and therfore displacement =t^3 +0.5t² +ct+ x (x is a made up constant variable). Where do i go now? – John Nov 26 '15 at 9:30
• Now put $d=12$ when $t=2$ and $d=34$ when $t=3$. So you have equations: $12=10+2c+x$ and $34=31.5+3c+x$. Now easily solve for the 2 constants and get the answers. Hope it helps..:-) – SchrodingersCat Nov 26 '15 at 10:02

Let me give you a hint for your second problem:

Again, the problem is to find the constant C after integrating from the acceleration to the velocity equation. To achieve this, you have to integrate the velocity equation again, to get the displacement equation. Now, you can use your knowledge for the displacement in the fourth second (calculate the difference of the displacement equation for t=4 and t=3) to get the c for the velocity equation. Then, simply solve it for t=5.

• Ok for problem number 2, so v=(3/2)t² +2t +c. But how do I find c? Because it says it travels 30m in the FOURTH second (this isnt an instantaneous quantity.) – John Nov 26 '15 at 9:25
• First find the displacement-equation by integrating for t again. Let this be disp(t). Then you know, that disp(4)-disp(3) = 30 and you can derive c from that – Ctx Nov 26 '15 at 9:27
• I integrated the velocity equation to get d(t)= 0.5x^3 +t² +ct+ x (x is a made up constant variable) but now i have three unknowns... – John Nov 26 '15 at 9:36
• Its disp(t) = 0.5t^3+t^2+ct+x. Now calculcate disp(4) - disp(3) = 30. What do you get as an equation? – Ctx Nov 26 '15 at 9:38
• Thanks you so much man, I finally worked it out. Problem one solved!! Cheers – John Nov 26 '15 at 9:43