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:
1309.1 Nm
Explanation:
Torque is given as a product of Moment of innertia and acceleration hence
T=Ia where T is torque and a is acceleration
To get acceleration, it is rate of change of speed per unit time hence where v and t represent velocity and time respectively while subscripts f and i represent final and initial respectively. Also, I is given by
where m js mass and r is radius hence the net torque can now be written as
By substituting the given figures then
Answer:
4. 10.0 m/s²
Explanation:
I) if initial velocity is 'v₀', the final velocity is 'v', the accelaration is 'a', the distance is 'L' and elapsed time if 't', then:
II) using these two equations after substitution v₀=0; v=30 and L=45: