Can anybody help me to find acceleration

[rashida564]

< Mentor Note -- poster has been reminded to use the Homework Help Template and show their work next time >[/color]

what is the acceleration for this equation Δx=4.0t+3.0t²
and thank you a lot

 

[cnh1995]

what is the acceleration for this equation Δx=4.0t+3.0t²
and thank you a lot

Please use proper homework template and show your attempt. It is mandatory when posting on homework help forums.

 

[rashida564]

and i also think the acceleration is equal to 6 m/s²

 

[rashida564]

https://img2.brain4.photobox.com/98...9970fa29ca2e4d6f4fe4513f848b27349dc6e249c.jpg

 

[rashida564]

 

[Chestermiller]

So, are you able to articulate your question?

 

[UchihaClan13]

One hint
What if you made the time interval infinitesimally small?
Wouldn't v=delta x/delta t (t tends to 0) become dx/dt
Do the same thing with the velocity
You should be able to get a=d^2x/dt^2
And yes you get v=4+6t
And a =dv/dt=6
Your answer is correct
UchihaClan13

 

[rashida564]

thank you UchihaClan13

 

[rashida564]

but can you me without calculus

 

[Chestermiller]

but can you me without calculus

If v₀ is the initial velocity (at time 0) and a is a constant acceleration, the velocity v at time t is $v=v_0+at$. The average velocity between time 0 and time t is $$v_{ave}=\frac{v_0+v}{2}=v_0+\frac{1}{2}at$$The distance traveled is equal to the average velocity times the time:
$$x=v_{ave}t=v_0t+\frac{1}{2}at^2$$

 

[rashida564]

then i should do that
1/2a=3
a=6

 

[Chestermiller]

then i should do that
1/2a=3
a=6

Yes. You were asking for a non-calculus derivation, so this is it.

 

[rashida564]

thank you