Respuesta :
Answer:
[tex]The\quad magnitude\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad \quad =2.52m/{ s }^{ 2 }\\ The\quad direction\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad =15.{ 72 }^{\o}\\The\quad car\quad begins\quad to\quad slide\quad out\quad \quad of\quad the\quad circle\quad after\quad 26.09s.\quad \quad \quad \quad[/tex]
Explanation:
Acceleration of a body is the change of velocity with respect to time.
- a) The magnitude of the car's acceleration is 2.52 m per second squared.
- b) The direction of the car's acceleration is at a angle of 15.72 degrees.
- c) The time taken by the car begin to slide out of the circle is 26.09 seconds.
What is acceleration?
Acceleration of a body is the change of velocity with respect to time.
Given information-
The mass of the car is 1900 kg.
Force applied forward is 1300 N.
The diameter of the circular path is 65 m.
The value of time is 13 s.
- a) The magnitude of the car's acceleration
The tangential acceleration is the ratio of force to the mass of the car is. Thus,
[tex]a_t=\dfrac{1300}{1900}\\a_t=0.684[/tex]
As the initial velocity of the car is zero. Thus by the linear motion of equation the velocity of the car can be given as,
[tex]v=u+at\\v=0+0.684\times13\\v=8.892[/tex]
The radial acceleration is the ratio of force to the mass of the car can be find out as,
[tex]ma_r=\dfrac{mv^2}{r}\\a_r=\dfrac{v^2}{\dfrac{d}{2}}\\a_r=\dfrac{(8.892)^2}{\dfrac{65}{2}}\\a_r=2.43\rm m/s^2[/tex]
Thus the resultant acceleration of tangential and radial acceleration is magnitude of the acceleration. Thus,
[tex]a_r=\sqrt{0.684\times2.43}\\a_r=2.52\rm m/s^2[/tex]
Thus the magnitude of the car's acceleration is 2.52 m per second squared.
- b) The direction of the car's acceleration
Direction of the ratio can be given as,
[tex]\theta=\tan^- \dfrac{a_t}{a_r}\\\theta=\tan^- \dfrac{0.684}{2.43}\\\theta=15.72^o[/tex]
Hence the direction of the car's acceleration is at a angle of 15.72 degrees.
- c) Time taken by the car begin to slide out of the circle-
The velocity can be given as,
[tex]v=\sqrt{\mu gh}[/tex]
Here, [tex]\mu[/tex] is the coefficient of static friction.
The time can be calculated by dividing the velocity by the acceleration as,
[tex]t=\dfrac{v}{a} \\t=\dfrac{\sqrt{1\times32.8\times32.5}}{0.684}\\t=26.09\rm s[/tex]
Thus the time taken by the car begin to slide out of the circle is 26.09 seconds.
Hence,
- a) The magnitude of the car's acceleration is 2.52 m per second squared.
- b) The direction of the car's acceleration is at a angle of 15.72 degrees.
- c) The time taken by the car begin to slide out of the circle is 26.09 seconds.
Learn more about the acceleration here;
https://brainly.com/question/11021097
