Suppose you increase your walking speed from 5m/s to 15m/s in a period of 1 s. What is your acceleration?

1 answer:

7 0

Acceleration=change in velocity divided by change in time or $a=\frac{\Delta v}{\Delta t}$

we are given that it goes from 5m/s to 15m/s
$\Delta v=v_f-v_0$
in this case, $\Delta v=15\frac{m}{s}-5\frac{m}{s}=10\frac{m}{s}$

and the change in time or $\Delta t$ is 1 second because this change occurs over a period of 1 second

so the acceleration is $a=\frac{10\frac{m}{s}}{1s}=10\frac{m}{s^2}$
the acceleration is $10\frac{m}{s^2}$

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A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s.

irakobra [83]

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system ( $\tau$ ), measured in Newton-meters, is:

$\tau = I\cdot \alpha$ (1)

Where:

$I$ - Moment of inertia, measured in Newton-meter-square seconds.

$\alpha$ - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

$\alpha = \frac{\omega - \omega_{o}}{t}$ (2)

Where:

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\omega$ - Final angular speed, measured in radians per second.

$t$ - Time, measured in seconds.

If we know that $\tau = 3\,N\cdot m$ , $\omega_{o} = 0\,\frac{rad}{s }$ , $\omega = 145.875\,\frac{rad}{s}$ and $t = 4\,s$ , then the moment of inertia of the motor is: