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adoni [48]
3 years ago
15

A student is running to catch the campus shuttle bus, which is stopped at the bus stop. The student is runnign at a constant spe

ed of 6.0 m/s; she cant run any faster. When trhe student is still 60 m from the bus, it starts to pull away, moving with a constant acceleration of 0.170m/s2. (a) For how much time and what distance does the student have to run at 5.0 m/s before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling?
Physics
1 answer:
Serggg [28]3 years ago
4 0

Answer:

Part a)

t = 16.8 s

d = 100.8 m

Part b)

v_f = 2.86 m/s

Explanation:

Part a)

Constant speed by which the student will run is given as

v = 5 m/s

now after some time if student is going to overtake the position of bus

so here the final positions will be same

so we have

x_{bus} = x_{student}

0 + \frac{1}{2}at^2 + d = v_{student} t

\frac{1}{2}(0.170)t^2 + 60 = 5 t

0.085 t^2 - 5t + 60 = 0

so it is

t = 16.8 s

So student will run the total distance

d = vt

d = (6)(16.8)

d = 100.8 m

Part b)

Speed of bus when student reach the bus is given as

v_f = v_i + at

v_f = 0 + (0.170)(16.8)

v_f = 2.86 m/s

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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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Answer:

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

b)n_1=7.735 \,rev

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

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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}
  • full speed of rotation, N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}
  • 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

∴ angular speed ω:

\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}

(c)

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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