Velocity Acceleration and Displacement

[Shanice80]

Homework Statement

Hi am a physics beginner love it but seem to have issues with understanding Linear motion .. Anyway i have a question ..will be glad if could help .
" A car moves along a road in such a way that the distance x ,traveled in a time t ,is given by x=a+bt(to the power 3) where A and B have constant values of 3.5m and 2.5ms-3 respectively .
Derive expressions for the velocity and acceleration as a function of time and calculate the values of displacement ,velocity and acceleration at t=2.0s .
Then draw a rough sketch to show how the velocity and acceleration vary with time..
Thanks ...

Homework Equations

The Attempt at a Solution

 

[rock.freak667]

ok so you know that x=A+Bt³ and you are given A and B, now write down the equation of x with A and B with their respective values.

Homework Equations

How do you find velocity when given displacement as a function of time? (i.e. what is the relation between velocity, displacement and time?

Similar for acceleration.


EDIT: Read up on http://en.wikipedia.org/wiki/Kinematics"

 

[tomcue]

To derive expressions for velocity and acceleration, we can use the basic equations for kinematics:

Velocity = change in displacement / change in time = (x2-x1) / (t2-t1)

Acceleration = change in velocity / change in time = (v2-v1) / (t2-t1)

We can also use the given equation x = a + bt^3 to find the velocity and acceleration as a function of time.

Taking the derivative of the equation, we get:

v = 3bt^2

a = 6bt

To calculate the values at t=2.0s, we substitute t=2.0s into the equations above.

For displacement, x = a + bt^3 = 3.5 + 2.5(2.0)^3 = 27.5m

For velocity, v = 3bt^2 = 3(2.5)(2.0)^2 = 15ms^-1

For acceleration, a = 6bt = 6(2.5)(2.0) = 30ms^-2

To show how the velocity and acceleration vary with time, we can plot a graph with time on the x-axis and velocity/acceleration on the y-axis. The graph will show a linear increase in velocity and a constant acceleration of 30ms^-2.

I hope this helps! Let me know if you have any other questions. Keep up the good work with physics!