Determining initial velocity and instant velocity

AI Thread Summary
The discussion focuses on calculating the initial velocity (v0x) of a particle with a given acceleration function, ax(t) = 2.90t - 2.06, to ensure it has the same x-coordinate at t = 4.06 as at t = 0. The equations for velocity and position are established, integrating the acceleration to find expressions for vx and x. The user struggles with incorporating the specific time values into the equations to find v0x. The solution involves equating the position expressions at t = 0 and t = 4.06 and solving for v0x. The thread emphasizes the importance of correctly applying integration and substitution in kinematic equations.
Norfonz
Messages
55
Reaction score
1

Homework Statement



The acceleration of a particle is given by ax(t) = 2.90t - 2.06

Q. Find the initial velocity v0x such that the particle will have the same x-coordinate at time t = 4.06 as it had at t = 0.

Homework Equations



vx = v0x + \intaxdt evaluated from 0 to t.
x = x0 + \intvxdt evaluated from 0 to t.

This is defined for straight-line motion with varying acceleration, which appears to fit my scenario.

The Attempt at a Solution



vx = v0x + \int(2.90t - 2.06)dt evaluated from 0 to t.
vx = v0x + 1.45t2 - 1.03t

x = x0 + \int(v0x + 1.45t2 - 1.03t)dt evaluated from 0 to t.
x = x0 + \int(1.45t2 - 1.03t)dt + v0x\intdt

x = x0 + v0xt + (1.45t3/3) - (1.03t2/2)

Unfortunately this is as far as I got. I don't understand how to incorporate the values I was given for t. Could someone point me in the right direction please?
 
Physics news on Phys.org
Norfonz said:
x = x0 + v0xt + (1.45t3/3) - (1.03t2/2)

Unfortunately this is as far as I got. I don't understand how to incorporate the values I was given for t. Could someone point me in the right direction please?

Well from where, when t=4.06, what is x equal to ? (in terms of v0x and x0)

And when t=0, what is x equal to ? (in terms of v0x and x0)


Then just put those equal to each other and solve for v0x
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Finding Initial Velocity v0x for Particle with Non-Constant Acceleration
Replies
2
Views
4K
Electron between two parallel plates
Replies
6
Views
3K
Projectile Motion Problem Involving Initial Velocity and Gravity Only
Replies
11
Views
1K
Solve 1-D Motion Problems: Homework Statement
Replies
8
Views
2K
Completely stumped on this one -- Kinematic Conceptual problem
Replies
34
Views
4K
Calculating Vx and Vy given V, X, Y
Replies
29
Views
9K
2 dimensional projectile motion with rifle and bullet
Replies
1
Views
6K
Finding Initial Velocity with Given Distance and Angle?
Replies
2
Views
1K
Solve 2D Motion Q: Wiley E Coyote & Roadrunner
Replies
1
Views
2K
Finding velocity and position of ##a(t)=−\omega^2(C_1\cos\theta+C_2\sin\theta)##
Replies
12
Views
1K

Hot Threads

Recent Insights

Back
Top