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Tresset [83]
3 years ago
11

A bicycle travels at a velocity of 2.33 m/s, and has a displacement of -58.3 m. How much time did it take?

Physics
1 answer:
Sedbober [7]3 years ago
4 0

Answer:

The time will be: 25.02 seconds

Explanation:

Given

A bicycle travels at a velocity of 2.33 m/s.

i.e.

  • v = 2.33 m/s

and

  • displacement = d = 58.3 m

As we know the formula to find the time

Time = Displacement/Velocity

        = d/v

        = 58.3/2.33

        = 25.02 seconds

Therefore, the time will be: 25.02 seconds

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A wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly wit
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A) +1.67 rad/s^2

The angular acceleration of the wheel is given by

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

where

\omega_i = -6.00 rad/s is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

\omega_f = 4.00 rad/s is the final angular velocity (anticlockwise, so with a positive sign)

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\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant t_1 where the angular velocity becomes positive (so, \omega_i=0) and the time corresponding to the final instant t_2 = 6.0 s, where \omega_f = +6.00 rad/s. We can find this time interval by using

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

And solving for \Delta t we find

\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant t_1=0 when \omega_i=-4.00 rad/s) and the time corresponding to the instant in which the velovity becomes positive t_2, when \omega_f = 0 rad/s. We can find this time interval by using

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

And solving for \Delta t we find

\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{0-(-4.00 rad/s)}{+1.67 rad/s^2}=2.4 s

D) 5.6 rad

The angular displacement of the wheel is given by the equation

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where we have

\omega_i = -6.00 rad/s is the initial angular velocity of the wheel

\omega_f = 4.00 rad/s is the final angular velocity

\alpha=+1.67 rad/s^2 is the angular acceleration

Solving for \theta,

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{((+6.00 rad/s)^2-(-4.00 rad/s)^2}{2(+1.67 rad/s^2)}=5.6 rad

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