So, the car takes approximately 8.43 seconds to travel 100 meters.

[tomc612]

Hi,
another question I am having trouble withView attachment 6215

so my thought at the moment is to integrate a(t), which results in 1/2at^2 which is the kinematic equation for distance and then solve for the equation to equal 100.

Just doesn't seem right to me and possibly too easy a solution.. think I am missing something here.

Any help appreciated

Thanks

 

[Prove It]

Hi,
another question I am having trouble with

so my thought at the moment is to integrate a(t), which results in 1/2at^2 which is the kinematic equation for distance and then solve for the equation to equal 100.

Just doesn't seem right to me and possibly too easy a solution.. think I am missing something here.

Any help appreciated

Thanks

Well for one thing, acceleration is the derivative of VELOCITY, not distance.

You will need to integrate TWICE and use some known conditions to evaluate the constants.

 

[tomc612]

hi,
anti derivative of a(t) = 1/2at^2

anti-derivative of 1/2at^2 = 1/6at^3

which doesn't look like any of the kinematic equations.

Still puzzled by this one as S= S(i) + V(i)+ 1/2at^2 which the derivative will give you a(t). Then solve with S = 100m

Any advice appreciated

Thanks

 

[HallsofIvy]

hi,
anti derivative of a(t) = 1/2at^2

No, it isn't. You were told that a(t)= t, not a(t)= at. The anti-derivative of t is $$v(t)= \frac{t^2}{2}+ C$$ for some constant C. Since v(0)= C, that constant is the speed at t= 0. And since, in this problem, the car is initially stopped, C= v(0)= 0.

anti-derivative of 1/2at^2 = 1/6at^3

Again, there is no constant "a". The anti-derivative of the velocity function, $$\frac{t^2}{2}+ C$$, is $$\frac{t^3}{6}+ Ct+ D$$ where D is another constant (here, the initial position).

which doesn't look like any of the kinematic equations.

The "kinematic equations" you are referring to assume constant acceleration which isn't the case here.

Still puzzled by this one as S= S(i) + V(i)+ 1/2at^2 which the derivative will give you a(t).

No, it won't. The second derivative will give you the constant a. There is no such a in this problem, the acceleration is not constant.

Then solve with S = 100m

That is assuming you took the initial position to be at S= 0.

Any advice appreciated

Thanks

My advice would be to start learning, and understanding, the material rather than memorizing formulas! You will always, as here, run into situations where the basic assumptions (that the acceleration is a constant) for those formulas do not hold.

 

[tomc612]

Hi,
yeah really struggling with this one.

So as its listed as a differential application, I've assumed there would be a First or Second Order differential to apply.

if a(t) = t just doesn't seem to be a lot to go on.

I can only think of integrating a(t) and then solving for t with the second integration as that would represent distance.

Any more help appreciated

Thanks

No, it isn't. You were told that a(t)= t, not a(t)= at. The anti-derivative of t is $$v(t)= \frac{t^2}{2}+ C$$ for some constant C. Since v(0)= C, that constant is the speed at t= 0. And since, in this problem, the car is initially stopped, C= v(0)= 0. Again, there is no constant "a". The anti-derivative of the velocity function, $$\frac{t^2}{2}+ C$$, is $$\frac{t^3}{6}+ Ct+ D$$ where D is another constant (here, the initial position). The "kinematic equations" you are referring to assume constant acceleration which isn't the case here. No, it won't. The second derivative will give you the constant a. There is no such a in this problem, the acceleration is not constant. That is assuming you took the initial position to be at S= 0. My advice would be to start learning, and understanding, the material rather than memorizing formulas! You will always, as here, run into situations where the basic assumptions (that the acceleration is a constant) for those formulas do not hold.

 

[MarkFL]

We are given:

\(\displaystyle a(t)=t,\,v_0=0,\,x_0=0\)

Now, since:

\(\displaystyle a(t)\equiv\d{v}{t}\)

We may write:

\(\displaystyle \d{v}{t}=t\)

So, we may now integrate, using the boundaries and switching dummy variables of integration:

\(\displaystyle \int_{0}^{v(t)}\,da=\int_0^t b\,db\)

Apply the FTOC:

\(\displaystyle v(t)=\frac{1}{2}t^2\)

Okay, now since:

\(\displaystyle v(t)\equiv\d{x}{t}\)

We may write:

\(\displaystyle \d{x}{t}=\frac{1}{2}t^2\)

So, we may now integrate, using the boundaries and switching dummy variables of integration:

\(\displaystyle \int_{0}^{x(t)}\,da=\frac{1}{2}\int_0^t b^2\,db\)

Apply the FTOC:

\(\displaystyle x(t)=\frac{1}{6}t^3\)

To find the time (in seconds) it takes the car to travel 100 m, we may set $x(t)=100$ and solve for $t$:

\(\displaystyle 100=\frac{1}{6}t^3\)

\(\displaystyle t=\sqrt[3]{600}=2\sqrt[3]{75}\approx8.43432665301749\)