Physics, 08.07.2020 01:01 woahhh17

A car moving at a speed of 25 m/s enters a curve that traces a circular quarter turn of radius 129 m. The driver gently applies the brakes, slowing the car with a constant tangential acceleration of magnitude 1.2 m/s2.a) Just before emerging from the turn, what is the magnitudeof the car's acceleration? b) At that same moment, what is the angle q between the velocity vector and theacceleration vector?
I am having trouble because this problem seems to have bothradial and tangential accleration. I tried finding the velocityusing V^2/R, but then that didnt take into account thedeceleration. Any help would be great.

Answers: 2

Show answers

Answers

Answer from: abdirisak33

8.7 m/s^2

82.15°

Explanation:

Given:-

- The initial speed of the car, vi = 25 m/s

- The radius of track, r = 129 m

- Car makes a circular " quarter turn "

- The constant tangential acceleration, at = 1.2 m/s^2

Solution:-

- We will solve the problem using rotational kinematics. Determine the initial angular velocity of car ( wi ) as follows:

$w_i = \frac{v_i}{r} \\\\w_i = \frac{25}{129}\\\\w_i = 0.19379 \frac{rad}{s}$

- Now use the constant tangential acceleration ( at ) and determine the constant angular acceleration ( α ) for the rotational motion as follows:

at = r*α

α = ( 1.2 / 129 )

α = 0.00930 rad/s^2

- We know that the angular displacement from the initial entry to the exit of the turn is quarter of a turn. The angular displacement would be ( θ = π/2 ).

- Now we will use the third rotational kinematic equation of motion to determine the angular velocity at the exit of the turn (wf) as follows:

$w_f^2 = w_i^2 + 2\alpha*theta\\\\w_f = \sqrt{0.19379^2 + 0.00930\pi } \\\\w_f = 0.25840 \frac{rad}{s}$

- We will use the evaluated final velocity ( wf ) and determine the corresponding velocity ( vf ) as follows:

$v_f = r*w_f\\\\v_f = 129*0.2584\\\\v_f = 33.33380 \frac{x}{y}$

- Now use the formulation to determine the centripetal acceleration ( ac ) at this point as follows:

$a_c = \frac{v_f^2}{r} \\\\a_c = \frac{33.3338^2}{129} \\\\a_c = 8.6135 \frac{m}{s^2}$

- To determine the magnitude of acceleration we will use find the resultant of the constant tangential acceleration ( at ) and the calculated centripetal acceleration at the exit of turn ( ac ) as follows:

$|a| = \sqrt{a^2_t + a_c^2} \\\\|a| = \sqrt{1.2^2 + 8.6135^2} \\\\|a| = 8.7 \frac{m}{s^2}$

- To determine the angle between the velocity vector and the acceleration vector. We need to recall that the velocity vector only has one component and always tangential to the curved path. Hence, the velocity vector is parallel to the tangential acceleration vector ( at ). We can use the tangential acceleration ( at ) component of acceleration ( a ) and the centripetal acceleration ( ac ) component of the acceleration and apply trigonometric ratio as follows:

$q = arctan \frac{a_c}{a_t} = arctan \frac{8.7}{1.2} \\\\q = 82.15 ^.$

The angle ( q ) between acceleration vector ( a ) and the velocity vector ( v ) at the exit of the turn is 82.15° .

Answer from: Quest

i guess the game developers wanted to give off a more gamey/cartoony feeling without trying to copy fortnite exactly

Answer from: Quest

the closer the planet is to the sun, the larger the gravity

Another question on Physics

Physics, 22.06.2019 02:10

Which statement correctly describes the relationship between frequency and wavelength?

Answers: 2

Answer

Physics, 22.06.2019 04:30

The image shows the positions of a car on a roller coaster track. arrange the cars in order based on their gravitational potential energy. begin with the lowest potential energy and end with the highest.

Answers: 1

Answer

Physics, 22.06.2019 16:30

In a classical model of the hydrogen atom, the electron moves around the proton in a circular orbit of radius 0.053 nm. what is the electron's orbital frequency? what is the effective current of the electron?

Answers: 3

Answer

Physics, 22.06.2019 19:30

Select light for the type of wave, adjust the wavelength so that the light is red, and increase the amplitude of the light to the max. then, select the start button at the source location to begin producing the waves. light is a form of electromagnetic wave, containing oscillating electric and magnetic fields. the wave amplitude detector mentioned above shows how the electric field oscillates in time at the location of the probe. the amplitude of the wave at the location of the probe is equal to the maximum electric field measured. how does the amplitude of the wave depend on the distance from the source?

Answers: 2

Answer

You know the right answer?

A car moving at a speed of 25 m/s enters a curve that traces a circular quarter turn of radius 129 m...

Questions