Derive an expression for the distance the rocket has traveled in terms of the initial and final speeds, and its acceleration. After you have this expression, plug in physical values from your problem to arrive at the distance the rocket has traveled
Displacement, Velocity and Acceleration
In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.
Linear Displacement
The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
I submitted the answer below. My professor asked me to answer his follow up question in bold 2. Derive an expression for the distance the rocket has traveled in terms of the initial and final speeds, and its acceleration. After you have this expression, plug in physical values from your problem to arrive at the distance the rocket has traveled. My thought process was to express the equation in terms of Distance = Velocity x Time. I was using the equation v = v0 + at to rearrange to fit in the distance equation, but when I plug in my known values I do not get 91m. Am I using the wrong equation?
Thanks
SOLVED PROBLEM
- A rocket blasts off and moves straight upward from the launch pad with constant acceleration. After 2.8 s the rocket is at a height of 91 m. (a)What are the magnitude and direction of the rocket’s acceleration? (b) What is its speed at this time?
1.Visualize the problem:
_____
I
I height, x = 91m
I
I
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2.Describe the problem in physics terms:
Known: Initial speed of rocket, vo= 0m/s
Initial height, x0 =0m
Height after 2.8s, x = 91m
Time takes for rocket to reach 91m, t = 2.8s
Find: the magnitude and the direction of rocket’s acceleration, a = ?
The speed of the rocket at 2.8s, v= ?
3.Plan a solution:
To find a solution, we will use 2 constant acceleration equations.
Equations: x = x0 + v0t + 1/2at2 (to find the rocket’s acceleration) We can use the constant acceleration of motion equation, x = x0 + v0t + 1/2at2, and convert to find a as we know x0, x, t and v0.
v = v0 + at (to find the rockets speed(velocity) since we will have calculated acceleration (a) with the above equation)
- Execute the plan
Calculations: Find a = ?
x = 0m +0m/s (2.8s) + a(2.8s)2/2
x = a(2.8s)2/2
2x = a(2.8s)2
2x/(2.8s)2 = a
Plug in x, 2(91m)/(2.8s)2= a
a = 182m/7.84s2 = 23 m/s2, is the acceleration and the rocket is moving in an upward direction
Find v = ?
v = v0 + at
v = 0m/s +23m/s2(2.8s)
v = 64 m/s, is the speed (velocity) at 2.8s
5.Check and evaluate
As constant acceleration is equal to average acceleration, we can plug in our values to the equation for average acceleration to confirm.
aav = V – v0 / t – t0
aav = 64 m/s – 0 m/s / 2.8s – 0s = 23m/s2
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