For a short time the position of a roller-coaster car along its path is defined by the equations r=25 m, θ=(0.3t) rad, and z=(−8 cosθ) m, where t is measured in seconds, Determine the magnitudes of the car's velocity and acceleration when t=4s .


Answer 1


Velocity = v = 2.24m/s

Acceleration = a = 0.20m/s²


Please see attachment below.


z=(−8 cosθ) and θ = 0.3t

z = -8Cos (0.3t)

V = dz/dt

a = v²/R.

Please see full solution below.

Answer 2

Final answer:

The roller coaster's velocity and acceleration at t=4 seconds is 7.64 m/s and 0.57 m/s² respectively.


The question is about understanding kinematics in cylindrical coordinates to analyze the motion of a roller coaster car. First, we need to understand that in polar coordinates, θ is changing with time t. So, the velocity vector v will have two components, one in the θ direction (rθ') and another in the z direction (z'). Given θ = 0.3t, we differentiate θ with respect to time to get θ' or dθ/dt = 0.3 rad/sec. Then, the z component of the velocity can be calculated by differentiating the equation of motion in the z-direction, z = -8 cos(θ), with respect to time. This gives z' = 8(0.3)sin(0.3t). So, at t=4s, z' = 8(0.3)sin(1.2) = 1.89 m/s. We then calculate rθ' = r*dθ/dt = 25*0.3 = 7.5 m/s.

The magnitude of velocity can then be calculated using the Pythagorean theorem: √((rθ')² + (z')²) = √((7.5)² + (1.89)²) = 7.64 m/s .

In a similar way, we can find the acceleration components. Given that r=25 m and is constant, radial acceleration is zero ( ar = r*(θ')²). The tangential acceleration is at = r*θ'' = r*d²θ/dt² =0 m/s² and z'' = dz'/dt = 8*0.3²*cos(0.3t). So, at t = 4s, z'' = 8(0.09)cos(1.2) = 0.57 m/s². The magnitude of the acceleration is given by √((ar)² + (at)² +(z'')²) = √((0)² + (0)² +(0.57)²)= 0.57 m/s².

Learn more about Motion in Cylindrical Coordinates here:


Related Questions

When you blow some air above a paper strip, the paper, rises. This is because 1.) the air above the paper moves faster and the pressure is lower 2.) the air above moves faster and the pressure is higher 3.) the air above the paper moves faster and the pressure remains constant 4.) the air above the paper moves slower and the pressure is higher 5.) the air above the paper moves slower and the pressure is lower
A cat leaps to catch a bird. If the cat's jump was at 60.0° off the ground and its initial velocity was point of its trajectory? 0.30 m 3.44 m/s, what is the highest O 13.76 m 0.45 m 0.90 m
In a charging process, 4 × 1013 electrons are removed from one small metal sphere and placed on a second identical sphere. Initially, both metal spheres were neutral. After the charging process, the electrical potential energy associated with the two spheres is found to be −0.063 J. What is the distance between the two spheres?
A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the ground. What one of the following statements is true if air resistance is neglected? The acceleration of the rock is zero when it is at the highest point. The speed of the rock is negative while it falls toward the ground. As the rock rises, its acceleration vector points upward. At the highest point the velocity is zero, the acceleration is directed downward. The velocity and acceleration of the rock always point in the same direction.
Write down the DE of motion of a particle moving under the influence of gravity and experiencing a resistive force. .

3. A particle of charge +7.5 µC is released from rest at the point x = 60 cm on an x-axis. The particle begins to move due to the presence of a charge ???? that remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved 40 cm if a) ???? = +20 µC and b) ???? = −20 µC?



HSBC keen vs kg get it yyyyyuuy




A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The curve may be regarded as an arc of a circle of radius 35.0 m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?



v = 14.35 m/s


As we know that crate is placed on rough bed

so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate

So here we have

\mu mg = (mv^2)/(R)

here we have

\mu g = (v^2)/(R)

now we know that

v = √(\mu Rg)

here we have

\mu = 0.600

R = 35 m

g = 9.81 m/s/s

now plug in all values in above equation

v = √((0.600)(35)(9.81))

v = 14.35 m/s

The electrons in the beam of a television tube have an energy of 19.0 keV. The tube is oriented so that the electrons move horizontally from north to south. At the electron's latitude the vertical component of the Earth's magnetic field points down with a magnitude of 42.3 μT. What is the direction of the force on the electrons due to this component of the magnetic field?



The direction is due south


From the question we are told that

     The energy of the electron is E = 19.0keV = 19.0 *10^3 eV

      The earths magnetic field is B = 42.3 \muT = 42.3 *10^(-6) T


Generally the force on the electron is perpendicular to the velocity of the elecrton and the magnetic field and this is mathematically reresented as

          \= F = q (\= v * \=B)

On the first uploaded image is an  illustration of the movement of the electron

    Looking at the diagram  we can see that in terms of direction  the magnetic force  is

             \= F  =q(\=v * \= B)= -( -\r i * - \r k)

                = -(- (\r i * \r k))

generally  i cross k = -j

      so the equation above becomes

             \= F = -(-(- \r j))

                = - \r j

This show that the direction is towards the south  


To calculate the change in kinetic energy, you must know the force as a function of _______. The work done by the force causes the kinetic energy change.'


Answer: The force is as a function of Distance


The force and distance must be parallel to each other. Only the component of the force in the same direction as the distance traveled does any work. Hence, if a force applied is perpendicular to the distance traveled, no work is done. The equation becomes force times distance times the cosine of the angle between them.

where both the force F and acceleration are vectors. This makes sense since both force and acceleration have a direction.

On the other hand, the kinetic energy


looks completely different. It doesn't seem to depend on the direction.




dW = F. dx

The volume control on a stereo is designed so that three clicks of the dial increase the output by 10 dB. How many clicks are required to increase the power output of the loudspeakers by a factor of 100?



300 clicks...


Output on 3 clicks = 10 dB

Increasing 10 by a factor of 100 equals 1000 dB so,

Its simple math, clicks will also increase in the same ratio and it shall take 300 clicks to increase the volume by a factor of 100.

An object of mass m moves to the right with a speed v. It collides head-on with an object of mass 3m moving with speed v/3 in the opposite direction. If the two objects stick together, what is the speed of the combined object, of mass 4m, after the collision


The speed of the combined object after collision is 0 m/s.

Total momentum before collision = total momentum after collision

m₁u₁ + m₂u₂ = (m₁ + m₂)a

m₁ = object 1 mass = m, u₁ = velocity of object 1 before collision = v, m₂ = mass of object 2 = 3m, u₂ = velocity of object 2 before collision = -v/3, a = velocity after collision

mv + 3m(-v/3) = (m + 3m)a

mv - mv = 4ma

0 = 4ma

a = 0 m/s

The speed of the combined object after collision is 0 m/s.

Find out more at:


the answer is 0 m/s


This question is describing the law of conservation of momentum

First object has mass =m

velocity of first object = v

second object = 3m

velocity of second object = v/3

the law of conservation of momentum is expressed as

m1V1 - m2V2 = (m1+ m2) V

substituting the parameters given;

making V as the subject of formular

V =(m_(1 ) V_(1) -m_(2)V_(2)  )/(m_(1)+m_(2)  )

V =

(mV - 3mv/3)/(m+ 3m)

V =(0)/(4m)

= 0 m/s

Other Questions