1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
bekas [8.4K]
3 years ago
9

The rocket's acceleration has components \(a_{x}(t)= \alpha t^{2}\) and \(a_{y}(t)= \beta - \gamma t\), where \(\alpha = 2.50 {\

rm m}/{\rm s}^{4}, \beta = 9.00 {\rm m}/{\rm s}^{2}\), and \(\gamma = 1.40 {\rm m}/{\rm s}^{3} \). At \(t = 0\) the rocket is at the origin and has velocity \({\vec{v}}_{0} = {v}_{0x} \hat{ i } + v_{0y} \hat{ j }\) with \(v_{0x} = 1.00 {\rm m}/{\rm s}\) and \(v_{0y} = 7.00 {\rm m}/{\rm s}\)
what is the max height??
Physics
1 answer:
lbvjy [14]3 years ago
5 0
 it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt 
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x} 
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y} 
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ] 
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt 
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases 
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume 
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ] 
You might be interested in
If youre walking from point a to b, the magnitude of your displacement will always be equal or less than or greater than your di
xenn [34]

The magnitude of your displacement can be equal to the distance you covered, or it can be less than the distance you covered.  But it can never be greater than the distance you covered.

This is because displacement is a straight line, whereas distance can be a straight line, a squiggly line, a zig-zag line, a line with loops in it, a line with a bunch of back-and-forths in it, or any other kind of line.

The straight line is always the shortest path between two points.

3 0
3 years ago
What does an unconformity in a sequence of rock layers reveal about geologic history?
andreev551 [17]
An unconformity, or abnormality in the layers of rocks, signifies that the process of layering was somehow interrupted. The only feasible choice for this would be (B), where erosion would have interrupted the layering process.
7 0
3 years ago
Read 2 more answers
A speedboat increases its speed from 14.5 m/s to 29.3 m/s in a distance of 172 m.
dalvyx [7]

Answer:

a. Acceleration, a = 1.88 m/s²

b. Time, t = 7.87 seconds.

Explanation:

Given the following data;

Initial velocity, U = 14.5m/s

Final velocity, V = 29.3m/s

Distance, S = 172m

a. To find the acceleration of the speedboat;

We would use the third equation of motion;

V² = U² + 2aS

Substituting into the formula

29.3² = 14.5² + 2a*172

858.49 = 210.25 + 344a

344a = 858.49 - 210.25

344a = 648.24

a = 648.24/344

Acceleration, a = 1.88 m/s²

b. To find the time;

We would use the first equation of motion;

V = U + at

29.3 = 14.5 + 1.88t

1.88t = 29.3 - 14.5

1.88t = 14.8

Time, t = 14.8/1.88

Time, t = 7.87 seconds.

6 0
3 years ago
A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant dece
seraphim [82]

Answer:

The value of acceleration that accomplishes this is 8.61 ft/s² .

Explanation:

Given;

maximum distance to be traveled by the car when the brake is applied, d = 450 ft

initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s

final velocity of the car when it stops, v = 0

Apply the following kinematic equation to solve for the deceleration of the car.

v² = u² + 2as

0 = 88.02² + (2 x 450)a

-900a = 7747.5204

a = -7747.5204 / 900

a = -8.61 ft/s²

|a| = 8.61 ft/s²

Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .

4 0
2 years ago
Two balls, each with a mass of 0.890 kg, exert a gravitational force of 8.06 × 10−11 n on each other. how far apart are the ball
Bond [772]
This problem involves Newton's universal law of gravitation and the equation to follow would be.

F = GM₁M₂/r²  

Given: M₁ = 0.890 Kg;  M₂ = 0.890 Kg;  F = 8.06 x 10⁻¹¹ N; G = 6.673 X 10⁻¹¹ N m²/Kg²

Solving for distance r = ?

r = √GM₁M₂/F

r = √(6.673 x 10⁻¹¹ N m₂/Kg²)(0.890 Kg)(0.890 Kg)/ 8.06 x 10⁻¹¹ N

r = 0.81 m 
6 0
3 years ago
Other questions:
  • What is question 22?
    6·2 answers
  • The total power input to the solar cell is 2.4W when the efficiency is 0.20.
    15·1 answer
  • A freight train leaves the train station 4hrs before a passenger train. The two trains are traveling in the same direction on pa
    7·1 answer
  • Which of the following occurs as light travels farther from its source?
    11·1 answer
  • How many moles of phosphorus trichloride are in 200 grams of the molecule???
    13·1 answer
  • A toy of mass 0.155 kg is undergoing simple harmonic motion (SHM) on the end of a horizontal spring with force constant 305 N/m
    13·1 answer
  • <
    14·1 answer
  • What kind of magnets will attract
    6·2 answers
  • Help me please<br><img src="https://tex.z-dn.net/?f=%20%5C%5C%20" id="TexFormula1" title=" \\ " alt=" \\ " align="absmiddle" cla
    12·2 answers
  • A block with a mass of 33.0 kg is pushed with a horizontal force of 150 N. The block moves at a constant speed across a level, r
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!