Ask Me Help Desk - a particle moves along a line with acceleration a(t)=5t+2. the velocity after 2 sec I

a particle moves along a line with acceleration a(t)=5t+2. the velocity after 2 sec I

I need how to calculate the initial velocity and the velocity after 4 seconds

Do you even understand what is meant by the given equation of a(t) = 5t+2?

Quote:

Originally Posted by Unknown008

Do you even understand what is meant by the given equation of a(t) = 5t+2?

Yed

Then, do explain what you understand.

Quote:

Originally Posted by Unknown008

Then, do explain what you understand.

Take d antiderivative then I'm lost

Okay... could you post the antiderivative that you obtained?

Quote:

Originally Posted by Unknown008

Okay... could you post the antiderivative that you obtained?

5/2t^2 + 2t

No, that's not completely it. You forgot the constant of integration.

You will need to use the additional given information that you got. Your question says that the velocity is a certain amount at 2 seconds.

Put back this constant of integration, and then substitute the values of t = 2 and v = whatever value you were given to get the value of the constant of integration.

Finally, could you explain what the equation you got means?

Quote:

Originally Posted by Unknown008

No, that's not completely it. You forgot the constant of integration.

You will need to use the additional given information that you got. Your question says that the velocity is a certain amount at 2 seconds.

Put back this constant of integration, and then substitute the values of t = 2 and v = whatever value you were given to get the value of the constant of integration.

Finally, could you explain what the equation you got means?

Idk 2 get the constant

Do you understand kinematics?

By that, I mean, how acceleration, velocity and displacement are interrelated and how one can obtain the other with reasonably simple steps?

Quote:

Originally Posted by Unknown008

Do you understand kinematics?

By that, I mean, how acceleration, velocity and displacement are interrelated and how one can obtain the other with reasonably simple steps?

No

Okay.

Do you know what each of them are?

In simple terms:
Displacement - Distance between the starting point of an object to the finish point of an object, with the direction specified.
Velocity - This is the speed of a moving object, in the specified direction. This is also the rate of change of displacement.
Acceleration - This is the rate of change of velocity of a moving object.

Basically:

http://p1cture.it/images/4cfb57196521a060f2eb.png

Now, I'll have to explain some basic integration...

Say, we have to find an equation for y, given that ${\frac{dy}{dx} = x^2 + x + 1}$ with respect to x, and are given that y = 2 when x = 0

${y = \int\ \frac{dy}{dx}\ dx = \int\ x^2 + x + 1\ dx = \frac13x^3 + \frac12x^2 + x + c}$

When x = 0, y = 2:

${2 = \frac13(0)^3 + \frac12(0)^2 + (0) + c}$

So, c = 2.

Therefore, the equation for y is:

${y = \frac13x^3 + \frac12x^2 + x + 2}$

Does that make sense?

Combining those two now.

a(t) = 5t + 2

Means that at any time t, the acceleration of the particle will be given by 5t + 2. For instance, at t = 0 (at the start), the acceleration is a(0) = 5(0) + 2 = 2 m/s^2

Similarly, at t = 10 s (10 s after the start), the acceleration is given by a(10) = 5(10) + 2 = 52 m/s^2

Now, if you integrate (inverse differentiate) the acceleration, you get the velocity. Find first the equation for the velocity of the particle. Then, you should be able to use what you were given to answer the questions asked to you.