Calculating Particle Position with Constant Acceleration

[Rybka]

Homework Statement

A particle is accelerated according to a(t) = 3.0t m/s^2. What is the position of the particle if it starts out with v=2.0 m/s and x = 0 at t = 0.

Homework Equations

s(t)=1/2at^2 + vot + xo

The Attempt at a Solution

I took the integral of a(t) to get my v(t), and then took the integral of v(t) to get my x(t). None of my answers match with the given choices.

The possible answers according to this sheet are:

(a). x(t) = 3.0t^2 + 2.0t m
(b). x(t) = 3.0t^3 + 2.0t m
(c). x(t) = t^3 +2.0t m

 

[BvU]

There being no difference whatsoever between answer a and answer b ?

By the way: what does your answer look like ?

 

[TSny]

Homework Equations

s(t)=1/2at^2 + vot + xo

Note that this equation is not relevant to this problem since the acceleration is not constant in this problem.

The Attempt at a Solution

I took the integral of a(t) to get my v(t), and then took the integral of v(t) to get my x(t).

That sounds good.

None of my answers match with the given choices.

The possible answers according to this sheet are:

(a). x(t) = 3.0t^2 + 2.0t m
(b). x(t) = 3.0t^2 + 2.0t m
(c). x(t) = t^3 +2.0t m

I agree that none of these answers is correct.

 

[Rybka]

There being no difference whatsoever between answer a and answer b ?

Yeah, sorry. I fixed that.

Nonetheless, the professor seems to have made a mistake on this problem set.

Thanks guys!

 

[HalfLostAndSearching]

I would like to clarify that the given options for the possible answers are incorrect. The correct answer for the position of the particle with constant acceleration can be found using the equation s(t) = 1/2at^2 + v0t + x0, where a is the acceleration, v0 is the initial velocity, and x0 is the initial position.

In this case, the particle starts with an initial velocity of 2.0 m/s and an initial position of 0 at t = 0. Therefore, plugging in the given values, we get:

s(t) = 1/2(3.0t^2) + (2.0)(t) + 0
= 1.5t^2 + 2.0t m

Hence, the correct answer is (d). x(t) = 1.5t^2 + 2.0t m. It is important to note that the unit for position is meters (m), not meters per second (m/s) as shown in the given options. This is because the integral of acceleration gives the change in velocity, and the integral of velocity gives the change in position.

It is also important to mention that the given options do not match with the equation provided in the statement, which is s(t) = 1/2at^2 + vot + xo. This may have caused confusion in finding the correct answer.

In conclusion, the correct answer for the position of the particle with constant acceleration is x(t) = 1.5t^2 + 2.0t m, and the options provided in the statement are incorrect.