Finding Time T with Given Velocity and Acceleration

[rjs123]

Homework Statement

An object is traveling in a straight line with an initial velocity of 3 m/s and an acceleration a(t) = st, where s = 2 m/s ^3 and t is measured in seconds. Find a time T such that v(T) = 21 m/s.

I wanted to use v = vo + at ... to find t...but the function of a(t) is kind of confusing...also i just started physics a couple weeks ago...so bear with me.

 

[SammyS]

Homework Statement

An object is traveling in a straight line with an initial velocity of 3 m/s and an acceleration a(t) = st, where s = 2 m/s ^3 and t is measured in seconds. Find a time T such that v(T) = 21 m/s.

I wanted to use v = vo + at ... to find t...but the function of a(t) is kind of confusing...also i just started physics a couple weeks ago...so bear with me.

v = vo + at is true is acceleration is constant.

In general a(t), a as a function of t is given by, $\displaystyle a(t)=\frac{d}{dt} v(t)$, acceleration is the time derivative of the velocity.

Therefore, velocity is the anti-derivative of the acceleration.

$$v(t)=\int a(t)\,dt$$
You will need to evaluate the constant of integration by considering the initial conditions.


If you're more comfortable treating it as a definite integral, then $\displaystyle v(t)=\int_{t_0}^{t} a(t)\,dt \,.$

 

[rjs123]

ok...where does the initial velocity get plugged into that equation though.

When you integrate don't you just get 1/2*at^2

I got caught up on the cubed part of the acceleration...because I am used to seeing it squared...like 9.8 m/s ^2.

 

[tiny-tim]

hi rjs123!

(try using the X² icon just above the Reply box )

When you integrate don't you just get 1/2*at^2

no

(if a is constant, and you integrate a twice, you get 1/2 at²)

you have dv/dt = st, so integrate to get v ​

 

[rjs123]

i understand it goes

x(t) = position

x'(t) = velocity

x''(t) = acceleration

to get the velocity from the acceleration just integrate...i understand that...but setting up the problem is the thing I'm having difficulty with even though it is something simple...the thing that makes me troubled is that the acceleration isn't constant.

 

[gneill]

to get the velocity from the acceleration just integrate...i understand that...but setting up the problem is the thing I'm having difficulty with even though it is something simple...the thing that makes me troubled is that the acceleration isn't constant.

That's why you want to integrate the acceleration; because it is not constant but a function of time.

You have an expression for a(t), and you want to integrate a(t). So...

 

[Superstring]

ok...where does the initial velocity get plugged into that equation though.

At time t=0, the velocity is 3 m/s. When you integrate the acceleration you get a new function of t plus a constant. What do you thing that the constant should be to satisfy the conditions?

When you integrate don't you just get 1/2*at^2

If a didn't change with time (it was constant), then when you integrate you would get $\frac{1}{2}at^2$, but your acceleration DOES change with time so you will end up with something different.

I got caught up on the cubed part of the acceleration...because I am used to seeing it squared...like 9.8 m/s ^2.

The 's' part isn't the acceleration - 'st' together is the acceleration. When you multiply m/s³ by seconds, you get m/s² which are the correct units for acceleration.

I think the 's' is what's throwing you off. Try forgetting about units for the time being - the acceleration function is really just a(t)=2t. That should look more familiar for integration.

 

[rjs123]

ok so...if you integrate a(t) = 2t...just becomes $$v(t) = t^2$$...

v = vo + at

would this be the correct approach?

$$21 m/s = 3 m/s + 2t(t)$$
$$18 m/s = 2t^2$$
$$9 m/s = t^2$$
$$3 seconds = t$$

 

[Superstring]

ok so...if you integrate a(t) = 2t...just becomes $$v(t) = t^2$$...

You forgot about the constant of integration (and units). It should read:

$$v(t) = \frac{1}{2} st^2+C$$

Now you need to figure out what C is equal to with the initial condition that v(0)=3 m/s

v = vo + at

would this be the correct approach?

You can't use that equation unless a is constant, which it is not. You ALREADY have an equation that tells you the velocity at any point in time (or, at least you will once you solve for C) - you just need to solve for t when v(t)=21.

 

[SammyS]

...

Therefore, velocity is the anti-derivative of the acceleration.

$$v(t)=\int a(t)\,dt$$
You will need to evaluate the constant of integration by considering the initial conditions.

If you're more comfortable treating it as a definite integral, then $\displaystyle v(t)=\int_{t_0}^{t} a(t)\,dt \,.$

I see there have been quite a few posts since my first reply. But you seem to have missed the whole point of what I stated in the quoted text above.

You have in the original problem statement that the acceleration is given by: a(t) = s·t .

Therefore, taking the integral (anti-derivative) gives: $\displaystyle v(t)=\int st\,dt$, where s is a constant, namely s = 2 m/s³.

Therefore, v(t) = (1/2)s t² + C .

At t=0, we have v(0) = 3 m/s. So, what is C ?

Using that value for C and plugging-in s = 2, should give you v(t) .

Then solve v(T) = 21 m/s for T .

 

[rjs123]

I see there have been quite a few posts since my first reply. But you seem to have missed the whole point of what I stated in the quoted text above.

You have in the original problem statement that the acceleration is given by: a(t) = s·t .

Therefore, taking the integral (anti-derivative) gives: $\displaystyle v(t)=\int st\,dt$, where s is a constant, namely s = 2 m/s³.

Therefore, v(t) = (1/2)s t² + C .

At t=0, we have v(0) = 3 m/s. So, what is C ?

Using that value for C and plugging-in s = 2, should give you v(t) .

Then solve v(T) = 21 m/s for T .

thanks i got it now...I just started 8 days ago...so this is all new to me.

c = 3

21 = t^2 + c
18 = t^2
t = 4.2 s

 

[Superstring]

thanks i got it now...I just started 8 days ago...so this is all new to me.

c = 3

21 = t^2 + c
18 = t^2
t = 4.2 s

Correct