Integrating to find velocity and position equations

AI Thread Summary
The particle's acceleration is given by a(t) = pt² - qt³, with initial conditions of zero velocity and position. The velocity function is derived by integrating the acceleration, resulting in v(t) = p(t³/3) - q(t⁴/4) after determining the constant of integration is zero. The position function is obtained by integrating the velocity, yielding x(t) = p(t⁴/12) - q(t⁵/20), also with the constant set to zero. Both functions have been verified against the initial conditions, confirming their correctness. The approach to finding velocity and position through integration is clearly outlined.
uchicago2012
Messages
74
Reaction score
0

Homework Statement


The acceleration of a certain particle is a function of time: a(t) = pt2-qt3, where p and q are constants. Initially, the velocity and position of the particle are zero.
(a) What is the velocity as a function of time?
(b) What is the position as a function of time?

The Attempt at a Solution


I was just wondering if these looked correct:
a) I integrated a(t) to get
v(t) = p(t3/3) - q(t4/4) + c
then subbed in v(0) = 0 to find c = 0
so v(t) = p(t3/3) - q(t4/4)

b) Then I integrated v(t) to get
x(t) = p(t4/12) - q(t5/20) + c
then subbed in x(0) = 0 to find c =0
so x(t) = p(t4/12) - q(t5/20)
 
Physics news on Phys.org
looks good to me
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »

Similar threads

How to find the forces in a particular framework of light rods
Replies
11
Views
2K
Find velocity-time equation from velocity-position equation
Replies
11
Views
2K
Confused about kinematic equations
Replies
20
Views
2K
Position and acceleration in simple harmonic motion given velocity
Replies
16
Views
1K
Two spring-coupled masses in an electric field (one of the masses is charged)
Replies
1
Views
915
Finding Velocity of Car Slowing Down: v_0-c*(t-t_1)^2/2
Replies
3
Views
1K
Irodov 1.23 confusion: Calculate distance traveled and time elapsed for this decelerating particle
Replies
2
Views
864
Can Ideal Gas Thermodynamics Explain This Circle Process?
Replies
1
Views
2K
Correct Derivation of the Adiabatic Condition? (PV Diagram)
Replies
4
Views
1K
Are the 2nd and 3rd problems in this video correctly solved? (charged dipole and organ pipe problems)
Replies
3
Views
1K

Hot Threads

Recent Insights

Back
Top