A rocket is fired at an angle from the top of a tower of height h 0 = 63.9m . Be

Question

A rocket is fired at an angle from the top of a tower of height h0 = 63.9m . Because of the design of the engines, its position coordinates are of the formx(t)=A+Bt2 and y(t)=C+Dt3, where A,B, C , and D are constants. Furthermore, the acceleration of the rocket 1.40s after firing is
a? =( 2.30 i^+ 2.30 j^)m/s2.
Take the origin of coordinates to be at the base of the tower.

1-Find the constant A

2-Find the constant B.

3-Find the constant C.

4- Find the constant D.

5-At the instant after the rocket is fired, what is its acceleration vector? (Enter your answers numerically separated by a comma.)

6- At the instant after the rocket is fired, what is its velocity?(Enter your answers numerically separated by a comma.)

Explanation / Answer

x(t) = A + Bt^2 and
y(t) = C + Dt^3

because they are vector equations, adding the x and y components together will give you an equation for the distance traveled by the rocket for a given time:

d(t) = A + Bt^2 + C + Dt^3

Differentiating this to get an equation for velocity:

dd/dx = v(t) = 2Bt(i) + 3Dt^2(j)

Note: the (i) and the (j) are there to represent the horizontal and vertical components of the velocity, and the constants (A) and (C) are not there because constants disappear when differentiated.

Differentiating this again to get an equation for acceleration:

a(t) = 2B(i) + 6Dt(j)

So this why there was calculus in this problem.

So to answer the questions:

1) Find A

If the origin is at the base of the tower, then the coordinates of the rocket when t=0 would be at the top of the tower: (0, 63.9). So because x=0 when t=0, it follows that:

x(t) = A + Bt^2 = 0

But because t=0, Bt^2 is also equal to 0. Therefore:

A=0

2) Find B

From the equation for acceleration:

a(t) = 2B(i) + 6Dt(j)

and also from the information that when t=1.40, a=2.30(i), it can be deduced that the horizontal component (2B) is equal to 2.30, regardless of the time. Therefore:

2B = 2.30
B = 1.15

3) Find C

From the equation:

y = C + Dt^3

And the knowledge that when t=0, y=63.9 (because the rocket is 63.9 m in the air, it can be deduced that:

Dt^3 = 0

Therefore:

C = 63.9

4) Find D

From the equation for acceleration:

a(t) = 2B(i) + 6Dt(j)

and also from the information that when t=1.40, a=2.30(j), it can be deduced that the vertical component (6Dt) is equal to 2.30. Therefore:

6Dt = 2.30
6(1.40)D = 2.3
D = 2.3 / 8.4
D = 0.27

5) Acceleration vector when t~0

Using:

a(t) = 2B(i) + 6Dt(j)

and given that (t) is almost equal to 0, 6Dt(j) will also be equal to 0. Therefore:

a(t) = 2B(i) + 0(j)

So the acceleration vector would be (2B, 0), or (2.30, 0).

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