Rocket Acceleration, position and velocity

[Lmaonade]

Homework Statement

"The acceleration of a certain rocket is given by a=Ct, where C is a constant. (a) Find the general position function x(t). (B) Find the position and velocity at t=5 s if x= 0 and v= 0 at t=0 and C= 3 m/s^3"

so

a= Ct

t=5s
x=0

v=0
t=0
C= 3 m/s^3

Homework Equations

The Attempt at a Solution

(a) Ill assume a "general position function" is just a small sketch of an "x vs t" Graph, since the rocket is accelerating the graph should just be a, concave curve up.

(B)I have no clue what exactly the question is asking for, with the given data.

 

[learningphysics]

velocity is the integral of acceleration. position is the integral of velocity. Use that for the first part.

 

[m2003]

I would approach this problem by first understanding the given information and what is being asked. From the given information, we know that the acceleration of the rocket is given by a=Ct, where C is a constant. This means that the acceleration is directly proportional to time.

To find the general position function x(t), we can use the basic kinematic equation x(t) = 1/2at^2 + v0t + x0, where a is the acceleration, t is time, v0 is the initial velocity, and x0 is the initial position. In this case, a=Ct, and we are given that v0=0 and x0=0. So the general position function becomes x(t) = 1/2Ct^2.

To find the position and velocity at t=5s, we can substitute t=5s into the general position function to get x(5) = 1/2C(5)^2 = 12.5C. This means that at t=5s, the rocket has traveled a distance of 12.5C.

To find the velocity at t=5s, we can take the derivative of the general position function with respect to time, which gives us v(t) = Ct. Substituting t=5s, we get v(5) = C(5) = 5C. This means that at t=5s, the velocity of the rocket is 5C.

In conclusion, the position at t=5s is 12.5C and the velocity at t=5s is 5C.