starting from rest , a formula one car accelerates uniformly at 25m\s2 for 30secs. what distance does it cover in the last one s
1 answer:
The distance covered in the last second of motion is 737.5 m
Explanation:
The motion of the car is a uniformly accelerated motion, so we can use the suvat equations.
First of all, we have to find the velocity of the car when the last second of motion starts, that is the velocity of the car after t = 29 s. We can use the equation:
v = u + at
where
u = 0 is the initial velocity
is the acceleration
Substituting t = 29 s,
Now we can find the distance covered in the last second of motion by using
where
u = 725 m/s is the velocity at the beginning of the last second
t = 1 s is the time interval considered
is the acceleration
Substituting,
Note that the acceleration of is not realistic for a car, but I still have used the data of the problem.
Learn more about acceleration:
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Answer:
a) , b)
Explanation:
The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:
Where is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:
Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:
a) t = 50 s.
b) t = 100 s.
Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:
Since , then the angular velocity is equal to zero. Therefore: