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3 years ago

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4. You are walking at a leisurely pace of .75 m/s when you hear the train coming and start running to catch it. Over the next 10

seconds, you speed up to 3.5 m/s and catch your train! What was your acceleration over that time?

1 answer:

hjlf3 years ago

5 0

Answer:

0.275 m/s²

Explanation:

The acceleration of an object is defined as the change in velocity over the change in time.

a = Δv/Δt
a = (v - v₀)/t

We are given the initial velocity, 0.75 m/s. The final velocity of the user in question is 3.5 m/s.

The time it takes for us to catch the train is 10 seconds. Using these variables, let's substitute into our equation.

a = (3.5 - 0.75)/10
a = 2.75/10
a = 0.275 m/s²

Our acceleration over that time was 0.275 m/s² in order to catch the train.

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A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

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a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

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e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

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time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

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$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

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using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

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