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3 years ago

Please help me with this very easy question I just don’t get it.

Physics

1 answer:

diamong [38]3 years ago

5 0

Answer:

150m

Explanation:

The relation of speed/time and distance/time is a derivative/integral one, as in speed is the derivative of distance (the faster you go, the faster the distance changes, duh!).

So we need to compute the integral of speed over time from 0.0s to 5.0s.

The easiest way here is to compute the area under the line (it's going to be faster than computing the acceleration and using a formula of distance based on acceleration).

The area under the line is a trapezoid with "height" 5s, and the bases 10m/s and 50m/s. Using the trapezoid area formula of h*(a + b)/2

distance = 5s * (10m/s + 50m/s) / 2 = 5s * 60m/s / 2 = 5s * 30m/s = 150m

Alternatively, we can use the acceleration formula:

a = (50m/s - 10m/s)/5s = 40m/s / 5s = 8m/s^2

distance = v0 * t + a * t^2 / 2 = 10m/s * 5s + 8m/s^2 * (5s)^2 / 2 = 50m + 8m * 25 / 2 = 50m + 100m = 150m.

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2 years ago

a sprinter with a mass of 80kg accelerates uniformly from 0 m/s to 9 m/s in 3 s. a.)what is the runners acceleration? b.) what i

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Part A:
Acceleration can be calculated by dividing the difference of the initial and final velocities by the given time. That is,
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Substituting,
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ANSWER: 3 m/s²

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ANSWER: 240 N 

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3 years ago

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4 years ago

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Answer:

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Solution:

As per the question:

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In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = $a^{2}$

Let the height of the bin be 'h'

Therefore the total area, $A_{t} = 4ah$

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C = 2sh

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Total cost, $C_{t} = 2a^{2} + 2ah$ (2)

From eqn (1):

$h = \frac{128}{a^{2}}$

Using the above value in eqn (1):

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$C(a) = 2a^{2} + \frac{256}{a}$

Differentiating the above eqn w.r.t 'a':

$C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}$

For the required solution equating the above eqn to zero:

$\frac{4a^{3} - 256}{a^{2}} = 0$

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$h = \frac{128}{4^{2}} = 8$

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4 years ago

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