Respuesta :
Answer: deceleration of [tex]1.904\ rad/s^2[/tex]
Explanation:
Given
Car is traveling at a speed of u=20 m/s
The diameter of the car is d=70 cm
It slows down to rest in 300 m
If the car rolls without slipping, then it must be experiencing pure rolling i.e. [tex]a=\alpha \cdot r[/tex]
Using the equation of motion
[tex]v^2-u^2=2as\\[/tex]
Insert [tex]v=0,u=20,s=300[/tex]
[tex]0-(20)^2=2\times a\times 300\\\\a=\dfrac{-400}{600}\\\\a=-\dfrac{2}{3}\ m/s^2[/tex]
Write acceleration as [tex]a=\alpha \cdot r[/tex]
[tex]-\dfrac{2}{3}=\alpha \times 0.35\\\\\alpha =-\dfrac{2}{1.05}\\\\\alpha =-1.904\ rad/s^2[/tex]
So, the car must be experiencing the deceleration of [tex]1.904\ rad/s^2[/tex].
The angular accelaration of a car traveling at 20.0 m/s on tires with a diameter of 70.0 cm, when car slows down to rest after traveling 300.0 m is 1.91 m/s
Angular acceleration: This can be defined as the rate of change of angular velocity. The s.i unit of angular acceleration is rad/s²
In other to solve the problem above, we will use the formula for calculating angular acceleration
a = αr.............. Equation 1
Where a = acceleration of the car's tires, α = angular acceleration of the car's tires, r = radius of the car's tires
Therefore,
α = a/r................. Equation 2
But, we need to calculate a using the equations of motion
v² = u²+2as................ Equation 3
Where v = final velocity = 0 m/s (rest), u = initial velocity= 20m/s, s = distance = 300 m
Substitute these values into equation 3
0² = 20²+2(300²)a
-400 = 600a
a = -400/600
a = -0.67 m/s²
going back to equation 2,
Given: r = 70/2 = 35 cm = 0.35 m
Substitute this values into equation 2,
α = -0.67/0.35
α = -1.91 rad/s²
Hence, the magnitude of the angular acceleration is 1.91 rad/s²
Learn more about angular acceleration here: https://brainly.com/question/20432894
