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emmainna [20.7K]

3 years ago

11

An electron has a constant acceleration of +3.3 m/s2. at a certain instant its velocity is +8.6 m/s. what is its velocity (a) 2.

3 s earlier and (b) 2.3 s later?

1 answer:

Citrus2011 [14]3 years ago

7 0

Hello!

The velocity function is:

$\mathsf{V = V_0 + at}\\ \\ \mathsf{V = 8.6 +3.3t}$

a) 2.3 s earlier, we can make t = -2.3 s, in relation to our initial time:

$\mathsf{V = 8.6 + 3.3\cdot(-2.3)}\\ \\ \mathsf{V = 8.8-7.59}\\ \\ \boxed{\mathsf{V = 1.21 \ m/s}}$

b) Now we can make t = 2.3 s:

$\mathsf{V = 8.6 + 3.3\cdot2.3}\\ \\ \mathsf{V = 8.6 + 7.59}\\ \\ \boxed{\mathsf{V = 15.59 \ m/s}}$

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$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

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

$r$ - Radius of rotation (Radius of a tire), measured in meters.

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$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

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

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

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

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$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

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