Why Does Projectile Motion Involve Zero X-Component Acceleration?

[jamiebean]

Homework Statement

The position of a particle is
4.9t^{3}\hat{i} +-4.4t^{4}\hat{j} m . what is the x component of the acceleration at time 3.4 s in unit of ms-2? t denotes time in second.

Relevant Equations

d=vi x t + 0.5at^2

I intended to finish the question with the equation of linear motion with constant acceleration, but it didn't work out. And I have no idea about the t^3 and t^4 of the position. How can I find the x component of the acceleration at time 3.4 s ? Where is the acceleration rate?

 

[PeroK]

Acceleration isn't constant. How, in general, do you find acceleration from a function of position?

 

[jamiebean]

Acceleration isn't constant. How, in general, do you find acceleration from a function of position?

a=dv/dt ??

 

[PeroK]

a=dv/dt ??

Of course!

 

[jamiebean]

Of course!

I'm quite new to differentiation so... I don't really know how to calculate arghhhh
a=dv/dt
I calculated dv=14.7t^2
but isn't it dt=d x 3.8=0?

 

[PeroK]

I'm quite new to differentiation so... I don't really know how to calculate arghhhh
a=dv/dt
I calculated dv=14.7t^2
but isn't it dt=d x 3.8=0?

I suggest you calculate the velocity and acceleration for this particle. If you don't know how to differentiate, then that might be a problem.

Paul's online maths is a good resource for all things calculus:

http://tutorial.math.lamar.edu/Classes/CalcI/DiffFormulas.aspx

 

[dio00]

Approach. You need to find dv/dt of the x-component. That takes a two-step process from x(t) -> a function of position. So to get to dv/dt, you need to differentiate dx/dt twice. What this means is first derive the velocity and then from velocity derive acceleration. Still unclear? No worries, look below.

The x-component is the first part of the equation -> 4.9t^3 (i.e x(t)=4.9t^3)

dx/dt = 14.7t^2 (which is also equal to v(t) = 14.7t^2 -> for the x-component)

Differentiate again [dv/dt]

dv/dt = 29.4t (which is also equal to a(t) = 29.4t -> for the x-component) [i.e instanteneous acceleration for x(t)]

Solution for (instantenous acceleration at t=3.4s) = 29.4 (3.4) = 99.96 m/s^2

If anything is unclear, I am happy to answer your questions.

 

[jamiebean]

Approach. You need to find dv/dt of the x-component. That takes a two-step process from x(t) -> a function of position. So to get to dv/dt, you need to differentiate dx/dt twice. What this means is first derive the velocity and then from velocity derive acceleration. Still unclear? No worries, look below.

The x-component is the first part of the equation -> 4.9t^3 (i.e x(t)=4.9t^3)

dx/dt = 14.7t^2 (which is also equal to v(t) = 14.7t^2 -> for the x-component)

Differentiate again [dv/dt]

dv/dt = 29.4t (which is also equal to a(t) = 29.4t -> for the x-component) [i.e instanteneous acceleration for x(t)]

Solution for (instantenous acceleration at t=3.4s) = 29.4 (3.4) = 99.96 m/s^2

If anything is unclear, I am happy to answer your questions.

thanks a lot :) I have got it! really really big thanksss

 

[jamiebean]

I suggest you calculate the velocity and acceleration for this particle. If you don't know how to differentiate, then that might be a problem.

Paul's online maths is a good resource for all things calculus:

http://tutorial.math.lamar.edu/Classes/CalcI/DiffFormulas.aspx

thanks a lot ! it really helped alotttttttt hahaha thank you!

 

[dio00]

thanks a lot :) I have got it! really really big thanksss

One other thing to note is that the x-component of a projectile motion almost always has an acceleration of 0, because the x-component does not contribute to the detour. This might be a little confusing, considering what I shared earlier. I thought it will be useful to know.