I need help with question 4

A particle moves along the x axis from the origin. The magnitude of the position vector at time t is

Maurinko [17]

1) The average velocity is $-2.1\cdot 10^5 m/s$

2) The instantaneous velocity is $64t-260t^3$

Explanation:

1)

The average velocity of an object is given by

$v=\frac{d}{t}$

where

d is the displacement

t is the time elapsed

In this problem, the position of the particle is given by the function

$x(t) = 32t^2 - 65t^4$

where t is the time.

The position of the particle at time t = 6 sec is

$x(6) = 32(6)^2 - 65(6)^4=-83,088 m$

While the position at time t = 12 sec is

$x(12)=32(12)^2-65(12)^4=-1,343,232 m$

So, the displacement is

$d=x(12)-x(6)=-1,343,232-(-83,088)=-1,260,144 m$

And therefore the average velocity is

$v=\frac{-1,260,144 m}{12 s- 6 s}=-2.1\cdot 10^5 m/s$

2)

The instantaneous velocity of a particle is given by the derivative of the position vector.

The position vector is

$x(t) = 32t^2 - 65t^4$

By differentiating with respect to t, we find the velocity vector:

$v(t) = x'(t) = 2\cdot 32 t - 4\cdot 65 t^3 = 64t - 260 t^3$

Therefore, the instantaaneous velocity at any time t can be found by substituting the value of t in this expression.

Learn more about velocity:

brainly.com/question/5248528

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6 0

3 years ago

If a stone dropped into a well reaches the water's surface after 3.0 seconds, how far did the stone drop before hitting the wate

pogonyaev

Let h = distance (m) to the water surface.

Initial velocity, u = 0 (because the stone was dropped).
Use the formula
h = ut + (1/2)gt^2
where g = 9.8 m/s^2 (acc. due to graity)
t = time (s)

h = (1/2)*(9.8)*(3^2) = 44.1 m

8 0

3 years ago

What is salinity ? ..

AveGali [126]

Answer:

<h3>Salinity is the saltiness or amount of salt dissolved in a body of water, called saline water. It is usually measured in g/L or g/kg. </h3>

7 0

2 years ago

Read 2 more answers

A spaceship approaches the earth with a speed 0.50c. A passenger in the spaceship measures his heartbeat as 9- beats per minute.

JulijaS [17]

Answer:

D) 61 beats per minute

Explanation:

5 0

3 years ago

A quarterback is set up to throw the football to a receiver who is running with a constant velocity v⃗ rv→rv_r_vec directly away

Artist 52 [7]

Answer:

a) V_o,y = 0.5*g*t_c

b) V_o,x = D/t_c - v_r

c) V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

d) Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

Explanation:

Given:

- The velocity of quarterback before the throw = v_r

- The initial distance of receiver = r

- The final distance of receiver = D

- The time taken to catch the throw = t_c

- x(0) = y(0) = 0

Find:

a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it. Express V_o,y in terms of t_c and g.

b) Find V_o,x, the initial horizontal component of velocity of the ball. Express your answer for V_o,x in terms of D, t_c, and v_r.

c) Find the speed V_o with which the quarterback must throw the ball.

Answer in terms of D, t_c, v_r, and g.

d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.

Solution:

- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:

y = y(0) + V_o,y*t_c - 0.5*g*t_c^2

0 = 0 + V_o,y*t_c - 0.5*g*t_c^2

V_o,y = 0.5*g*t_c

- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:

- We know that V_i, x = V_o,x + v_r. Hence,

x = x(0) + V_i,x*t_c

D = 0 + V_i,x*t_c

V_o,x + v_r = D/t_c

V_o,x = D/t_c - v_r

- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:

V_o = sqrt ( V_o,x^2 + V_o,y^2)

V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

- The angle with which it should be thrown can be evaluated by trigonometric relation:

tan(Q) = ( V_o,y / V_o,x )

tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )

Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

6 0

3 years ago