Question

A faulty model rocket moves in the xy-plane (the positive y direction is vertically upward). The rocket’s acceleration has components ax(t) = αt2 and ay(t) = β −γt, 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 ~v0 = v0x ˆi + v0y ˆj with v0x = 1.00 m/s and v0y = 7.00 m/s. (a) Calculate the velocity and position vectors as functions of time. (b) What is the maximum height reached by the rocket? (c) Sketch the path of the rocket. (d) What is the horizontal displacement of the rocket when it returns to y = 0?

Answer 1

Is a set of elements of a vector space that a given Linear transformation maps to zero

Answer 2

Answer:

part a)

velocity is given as

$\vec v = ( 1 + \frac{\alpha t^3}{3})\hat i + (7 + \beta t - \frac{\gamma t^2}{2})\hat j$

position is given as

$\vec r = (t +\frac{ \alpha t^4}{12})\hat i + (7t +\frac{ \beta t^2}{2} - \frac{\gamma t^3}{6})\hat j$

Part b)

maximum height is given as

$y_{max} = 341.4 m$

Part c)

Path will be the curved path

Part d)

displacement in x direction will be

$x = 39328 m$

Explanation:

$a_x = \alpha t^2$

$a_y  = \beta - \gamma t$

also we know the initial velocity as

$v_0 = 1\hat i + 7 \hat j$

part a)

from acceleration in x direction we can say

$a_x = \frac{dv_x}{dt} = \alpha t^2$

$\int dv_x = \int \alpha t^2 dt$

$v_x - 1 = \frac{\alpha t^3}{3}$

$v_x = 1 + \frac{\alpha t^3}{3}$

$v_x = 1 + 0.83 t^3$

now again integrate both sides

$\frac{dx}{dt} = (1 + 0.83 t^3)$

$\int dx = \int(1 + 0.83 t^3) dt$

$x = t +\frac{ \alpha t^4}{12}$

$x = t + 0.21 t^4$

now similarly for y direction

$a_y = \frac{dv_y}{dt} = \beta - \gamma t$

$\int dv_y = \int (\beta - \gamma t) dt$

$v_y - 7 = \beta t - \frac{\gamma t^2}{2}$

$v_y = 7 + \beta t - \frac{\gamma t^2}{2}$

$v_y = 7 + 9 t - 0.70 t^2$

now again integrate both sides

$\frac{dy}{dt} = (7 + 9t - 0.70t^2)$

$\int dy = \int(7 + 9t - 0.70t^2) dt$

$y = 7t +\frac{ \beta t^2}{2} - \frac{\gamma t^3}{6}$

$y = 7t + 4.5 t^2 - 0.233 t^3$

PART B)

For maximum height we know that velocity in y direction must be zero

so we will have

$v_y = 0 = 7 + 9t - 0.70 t^2$

by solving above we have

$t = 13.6 s$

now at this time the height is given as

$y = 7t + 4.5 t^2 - 0.233 t^3$

$y_{max} = 7(13.6) + 4.5(13.6)^2 - 0.233(13.6)^3$

$y_{max} = 341.4 m$

PART C)

path of the rocket will be curved path in xy plane

PART D)

when y = 0 then we have

$y = 0 = 7t + 4.5 t^2 - 0.233 t^3$

by solving above equation we have

$t = 20.8 s$

now at this time position of X is given as

$x = t + 0.21 t^4$

$x = 20.8 + 0.21(20.8)^4$

$x = 39328 m$