a) The tangential acceleration is [tex]0.6 m/s^2[/tex]
b) The total acceleration is [tex]1 m/s^2[/tex] at [tex]53.1^{\circ}[/tex] from the tangential direction
Explanation:
a)
The tangential acceleration of an object in circular motion is the acceleration of the object in the direction tangent to the circular trajectory.
In this problem, the car is moving along the circular road of radius
r = 20 m
The tangential acceleration (magnitude) is given by
[tex]a_t = \frac{\Delta v}{\Delta t}[/tex]
where [tex]\Delta v[/tex] is the change in linear speed in a time interval [tex]\Delta t[/tex].
In this problem, we are told that the linear speed increases at a rate of
[tex]0.6 m/s^2[/tex]
Therefore, the tangential acceleration is exactly equal to this value:
[tex]a_t = 0.6 m/s^2[/tex]
b)
In order to find the total acceleration, we have to compute the centripetal acceleration first (in the radial direction), which is given by
[tex]a_c = \frac{v^2}{r}[/tex]
where
v = 4 m/s is the instantaneous linear speed
r = 20 m is the radius of the road
Substituting,
[tex]a_c = \frac{4^2}{20}=0.8 m/s^2[/tex]
Now we can find the magnitude of the total acceleration, given by:
[tex]a=\sqrt{a_t^2+a_c^2}=\sqrt{0.6^2+0.8^2}=1 m/s^2[/tex]
while the direction is given by:
[tex]\theta=tan^{-1}(\frac{a_c}{a_t})=tan^{-1}(\frac{0.8}{0.6})=53.1^{\circ}[/tex]
measured from the direction tangent to the circle.
Learn more about circular motion:
brainly.com/question/2562955
brainly.com/question/6372960
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