Answer
Time period T = 1.50 s
time t = 40 s
r = 6.2 m
a)
Angular speed ω = 2π/T
= 
= 4.189 rad/s
Angular acceleration α = 
= 
= 0.105 rad/s²
Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²
b)The maximum speed.
v = 2πr/T
= 
= 25.97 m/s
So centripetal acceleration.
a = 
= 
= 108.781 m/s^2
= 11.1 g
in combination with the gravitation acceleration.
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Answer: Option B.
Since here the truck is moving on a circular track, it will experience centripetal force.
F(centripetal) = m × acc
or
where r is the radius of the track.
m is the mass of truck
v is the speed of the truck.
Given: v = <span>13 m/s
m = </span><span>1,600 kg
</span>F = 3300 Newton
To find = radius of track=?
r = 81.94 m
Therefore, radius of track is 81.94 m
When the car comes to a stop, the final velocity must be 0 m/s.
Since the car js decelerating in a forward direction, acceleration must be negative.
final v = initial v + a•t
0 = 20 + (-6)t
t = 3.33s
