Question 1: A truck on a straight road start from rest, accelerating at 2.0 m/s^

Question

Question 1:

A truck on a straight road start from rest, accelerating at 2.0 m/s^2 until it reaches a speed of 19.0 m/s. Then the truck travels for 28.0s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.0 s.

How long is the truck in motion?

What is the average velocity of the truck for the motion described?

Question 2:

As shown in the figure, three forces vectors act on an object. The magnitude of the forces as shown in the figure are F1 = 80N and is 30 degrees clockwise below the x-axis (-30 degrees), F2 = 60N points North (90 degrees), and F3 = 40N and points west (180 degrees).

Determine the x and y components of F1.

Determine the x and y components of F2.

Determine the x and y components of F3.

Determine the x and y components of the sum of these three forces.

Determine the magnitude and direction of the sum of these three forces.

Question 3:

A particle starts from the origin at t = 0 with a velocity of 6.0 i m/s and moves in the xy plane with a constant acceleration of (-2.0i + 4.0j) m/s^2.

When does, the particle achieve its maximum positive x coordinate?

How far is it from the origin at that instant?

Question 4:

A child throws a ball with an initial speed of 8.0 m/s at an angle of 40.0 degrees above the horizontal. The ball leaves her hand 1.0 m above the ground and experience negligible air resistance.

What is the magnitude of the ball’s velocity just before it hits the ground?

At what angle below the horizontal does the ball approach the ground?

Question 5:

A satellite is in a circular orbit 600 km above the Earth’s surface. The acceleration of gravity is 8.21 m/s^2 at this altitude. The radius of Earth is 6400 km. determine the speed of the satellite, and the time to complete one orbit around the Earth.

Explanation / Answer

Question 1)

inital velocity v0x = 0

acceleration ax = 2 m/s^2

final speed vx = 19 m/s

time t1 = (vx-v0x)/ax = 19/2 = 9.5 s

distance x1 = (1/2)*ax*t1^2 = (1/2)*2*9.5^2 = 90.25 m

for second part

time = t2 = 28 s

distacne x2 = vx*t2 = 19*28 = 532 m

for third part

t3 = 5 s

x3 = (vfx+vx)/2*t3

x3 = 19/2*5 = 47.5 m

total time T = 9.5 + 28 + 5 = 42.5 s

average velocity = (x1 + x2 +x3)/T = (90.25+532+47.5)/42.5 = 15.8 m/s

========================

Q2)

F1x = 80*cos30 = 69.3 N

F1y = -80*sin30 = -40 N

F2x = 60*cos90 = 0

F2y = 60*sin90 = 60 N

F3x = 40*cos180 = -40 N

F3y = 40*sin180 = 0

Fx = F1x + F2x + F3x = 69.3 + 0 - 40 = 29.3 N

Fy = F1y + F2y + F3y = -40 + 60 = 20 N

magnitude = sqrt(Fx^2+Fy^2) = sqrt(29.3^2+20^2) = 35.5

direction = tan^-1(Fy/Fx) = 34.3 anticlockwise above x axis

=========================

Q3)

the particle achieve its maximum positive x coordinate when v = 0

vx = v0x + ax*t

0 = 6 - (2*t)

t = 3 s

x = v0x*t + (1/2)*ax*t^2

x = (6*3) - ((1/2)*2*3^2)

x = 9 m

--------------

Q4)

along vertical

vy^2 - v0y^2 = 2*ay*dy

vy^2 - (8*sin40)^2 = -2*9.8*(0-1)

vy = 6.78 m/s

angle = tan^-1(vy/vx) = tan^-1(6.78/(8*cos40)) = 48 degrees

============

Q5)

orbitla speed

v = sqrt(gr) = sqrt(g*(R+h))

v = sqrt(8.21*(6400000+600000))

v = 7580.8 m/s

Time T = 2*pi*r/v

T = 2*pi*(6400000+600000)/7580.8

T = 5801.8 s

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