Question

A faulty model rocket moves in the xy-plane (the positive y-direction is vertically upward). The rocket's...

A faulty model rocket moves in the xy-plane (the positive y-direction is vertically upward). The rocket's acceleration has components ax(t)=αt2and ay(t)=βγt, where α = 2.50 m/s4, β = 9.00 m/s2, and γ = 1.40 m/s3. At t=0 the rocket is at the origin and has velocity v⃗ 0=v0xi^+v0yj^with v0x = 1.00 m/s and v0y = 7.00 m/s.

A. Calculate the velocity vector as a function of time.

Express your answer in terms of v0x, v0y, β, γ, and α. Write the vector v⃗ (t) in the form v(t)x, v(t)y, where the x and y components are separated by a comma.

B. Calculate the position vector as a function of time.

Express your answer in terms of v0x, v0y, β, γ, and α. Write the vector r(t)→ in the form r(t)x, r(t)y where the x and y components are separated by a comma.

D. Sketch the graph

E. What is the horizontal displacement of the rocket when it returns to

y=0?

Homework Answers

Answer #1

a(x,y) = alpha t^2 i + (beta - gamma)t j

we know that

a = vd/dt => dv = a dt

dv = [2.5 t^2 i + (9 - 1.4)t j] dt

dv = (alpha t^2 i + (beta - gamma t) j

integrating this we get

v = (alpha t^3/3 + v0x )i + (beta t - gamma t^2/2 + voy) j

V = 2.5/3 t^3 + v0x + 9 t - 1.4/2 t^2 + v0y

Vx,Vy = (alpha t^3/3 + v0x ),(beta t - gamma t^2/2 + voy)

V = (0.833 t^3 + 1)i + (9 t - 0.7t^2 + 7) j m/s

B)v = dr/dt => dr = v dt

dr = [(alpha t^3/3 + v0x )i + (beta t - gamma t^2/2 + voy) j]

r = (alpha t^4/12 + v0x t + r0x)i + beta t^2/2 - gamma t^3/6 + voy t + r0y) j

rx,ry = (alpha t^4/12 + v0x t + r0x), beta t^2/2 - gamma t^3/6 + voy t + r0y)

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A rocket is fired at an angle from the top of a tower of height h0...
A rocket is fired at an angle from the top of a tower of height h0 = 61.2 m . Because of the design of the engines, its position coordinates are of the form x(t)=A+Bt2 and y(t)=C+Dt3, where A, B, C , and D are constants. Furthermore, the acceleration of the rocket 1.20 s after firing is a=( 5.10 i+ 4.80 j)m/s2. Take the origin of coordinates to be at the base of the tower. a) Find the constant A....
The coordinates of an object moving in the xy plane vary with time according to the...
The coordinates of an object moving in the xy plane vary with time according to the equations x 5 25.00 sin vt and y 5 4.00 2 5.00 cos vt, where v is a constant, x and y are in meters, and t is in seconds. (a) Determine the components of velocity of the object at t 5 0. (b) Determine the components of acceleration of the object at t 5 0. (c) Write expressions for the position vector, the...
A soccer ball is kicked in such a manner that the x and y components of...
A soccer ball is kicked in such a manner that the x and y components of its position at any time are given by x = (19.6 m/s)t and y = (14.5 m/s)t − (4.90 m/s2)t2, where x and y will be in meters when t is in seconds. (a) Determine expressions for the x and y components of its velocity at any time. (Use the following as necessary: t.) vx =   m/s vy =   m/s (b) Determine expressions for...
A particle moves in the xy plane, starting from the origin at t=0 with an initial...
A particle moves in the xy plane, starting from the origin at t=0 with an initial velocity having an x-component of 6 m/s and y component of 5 m/s. The particle experiences an acceleration in the x-direction, given by ax=4t m/s2. Determine the acceleration vector at any later time. Determine the total velocity vector at any later time Calculate the velocity and speed of the particle at t=5.0 s, and the angle the velocity vector makes with the x-axis. Determine...
A particle moving in the xy-plane has velocity v⃗ =(2ti+(3−t2)j)m/s, where t is in s. What...
A particle moving in the xy-plane has velocity v⃗ =(2ti+(3−t2)j)m/s, where t is in s. What is the x component of the particle's acceleration vector at t = 7 s? What is the y component of the particle's acceleration vector at t = 7 s?
A 2400-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass)...
A 2400-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by v(t)=At+Bt2, where A and B are constants and time is measured from the instant the fuel is ignited. At the instant of ignition, the rocket has an upward acceleration of 1.90 m/s2 and 1.20 s later an upward velocity of 1.50 m/s . Part A) Determine A....
The acceleration of a particle moving only on a horizontal xy plane is given by a→=3ti^+4tj^,...
The acceleration of a particle moving only on a horizontal xy plane is given by a→=3ti^+4tj^, where a→ is in meters per second-squared and t is in seconds. At t = 0, the position vector r→=(19.0m)i^+(44.0m)j^ locates the particle, which then has the velocity vector v→=(5.40m/s)i^+(1.70m/s)j^. At t = 4.10 s, what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the x axis?
An object moves with constant speed in the x-direction, but in the y-direction it is subject...
An object moves with constant speed in the x-direction, but in the y-direction it is subject to an acceleration that increases linearly with time: a(t)=bt, where b is a constant. Assume there is no gravity. Derive an equation analogous to y=xtanθ−(g/2V^2cos^2θ0)x^2 giving the object's trajectory in this situation. Note that θ0θ0 is the angle measured from the horizontal, at which the projectile is launched, and v0v0 is its initial speed. Assume the object starts moving from the origin. Express your...
A student stands at the edge of a cliff and throws a stone horizontally over the...
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of v0 = 17.5 m/s. The cliff is h = 26.0 m above a flat, horizontal beach as shown in the figure. A student stands on the edge of a cliff with his hand a height h above a flat stretch of ground below the clifftop. The +x-axis extends to the right along the ground and the +y-axis extends up...
A positive charge q = 3.20 10-19 C moves with a velocity vector v = (2.00...
A positive charge q = 3.20 10-19 C moves with a velocity vector v = (2.00 x̂ + 3.00 y hat - 1.20 ẑ) m/s through a region where both a uniform magnetic field and a uniform electric field exist. (a) Calculate the total force on the moving charge (in unit-vector notation) taking vector B = (2.00 x̂ + 4.00 y hat + 0.900 ẑ) T and vector E = (4.20 x̂ - y hat - 2.00 ẑ) V/m. Fx...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT