A car with a velocity of 22 m/s is accelerated at a rate of 1.6 for 6.8s has the final velocity t be 32.88 m/s.
The acceleration means the amount of velocity changing per unit time.
The given data:
initial velocity, u = 22 m/s
time, t = 6.8 s
acceleration, a = 1.6
We will be using the equation of motion:
v = u + at
The final velocity become 32.88 m/s.
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the solution for the oscillatory movement allows to find the result for the amplitude of the initial displacement is:
The oscillatory periodic motion of a system occurs when there is a recovered force, in the special case that this force is proportional to the displacement is called simple harmonic motion, which is described by the expression.
x = A cos (wt + Ф)
w² = k/m
where x is the displacement, A the amplitude, w the angular velocity, t the time, k the spring constant, m the mass, and Ф a phase constant determined by the initial conditions.
Let's find the angular velocity/
w=
w = 25.8 rad/s
Let's look for the constant Ф, as the system is released from rest its initial velocity is zero, for zero time. The definition of speed is:
v=
v= - A w sin (wt +Ф)
0 = -A w sin Ф
Ф= 0
They indicate that at a given instant of the time the velocity is v= 50.0 cm/s and it is in a position x= 4.00 cm, let us write the equations for this time
Position.
4.00 = A cos 25.8t
Speed.
50.0 = - At 25.8 sin 25.8t
To solve the system, ;et's square and add.
Cos² 25.8t =
sin² 25.8t =
1 =
A =
A= 4.44 cm
In conclusion using the solution for the oscillatory movement we can find the result for the amplitude of the initial displacement is:
The range of motion is: A = 444 cm
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