Physics, 28.02.2020 03:56 322251

A rocket is fired from rest at x = 0 and travels along a parabolic trajectory described by y 2 = [120(10 3 )x] m. If the x component of acceleration is a x = ( 4 1 t 2 ) , m/s 2 , where t is in seconds, determine the magnitude of the rockets velocity and acceleration when t = 10 s.

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Answer from: jessicaisbaehood

The velocity is 1003.5 m/s and the acceleration is 103.1 m/ $s^{2}$ .

Explanation:

We need to find the parameter equation of x. To find it, we will need to integrate the x-component acceleration equation given to us, twice. Acceleration is:

a = $\frac{dv}{dt}$

dv = adt

∫dv = ∫adt

$\int\limits^v_0 \, dv$ = $\int\limits^t_0 {\frac{1}{4}t^{2} } \, dt$

v = $\frac{1}{12} t^{3} m/s$ .

Velocity is:

v = $\frac{dx}{dt}$

dx=vdt

Again, integrate both sides:

∫dx = ∫vdt

$\int\limits^x_0 dx$ = $\int\limits^t_0 {\frac{1}{12} t^{3} } \, dt$

x = $\frac{1}{48} t^{4} m$

Substitute our x equation into our parameter equation of y.

y $^{2}$ = [120( $10^{3}$ )x] m

y $^{2}$ = [120( $10^{3}$ )( $\frac{1}{48}$ )( $t^{4}$ ) ]

(take the square root of both sides and simplify)

y = 50 $t^{2}$

Now that we can represent our equation with respect to time, we can take the derivative to figure out the velocity. Remember that taking the derivative of a position function gives us the velocity function.

y = 50 $t^{2}$

vy= y = 100t

Let us write down the two equations for velocity we found:

vx = $\frac{1}{12} t^{3} m/s$

vy = 100t m/s

At t = 10 s:

vx = $\frac{1}{12} 10^{3}$ = 83.3 m/s

vy = 100(10)=1000 m/s

The magnitude of velocity is:

v = $\sqrt{(vx)^{2}+(vy)^{2} }$

v = $\sqrt{(83.3)^{2}+(1000)^{2} } = 1003.5 m/s$

To figure out the acceleration, we need to figure out ay which can be found by taking the derivative of the vy equation,

vy = 100t m/s

ay = vy = 100 m/ $s^{2}$

Since ax is given to us in the question, we have the following:

ax = $(\frac{1}{4}t^{2}) m/s^{2}$

ay = 100 m/ $s^{2}$

At t = 10 s:

ax = $(\frac{1}{4}10^{2})} =25 m/s^{2$

ay = 100 m/ $s^{2}$

The magnitude of acceleration is equal to:

a = $\sqrt{(ax)^{2}+(ay)^{2} }$

a = $\sqrt{(25)^{2}+(100)^{2} }$

a= 103.1 m/ $s^{2}$

Final Answers:

v = 1003.5 m/s

a= 103.1 m/ $s^{2}$

Answer from: Quest

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

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A rocket is fired from rest at x = 0 and travels along a parabolic trajectory described by y 2 = [12...

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