Question #5ced9

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Question #5ced9

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Answer 1 · 2017-10-22T15:28:38

(a) 0.6 s

(b) 1.2 s

Explanation:

As the fish falls with constant acceleration i.e $g$ $g$ , we can apply the equations of motion for both the questions.

(a)

$u = 6 m . s^{- 1}$ $u=6 m.s^(-1)$

According to question,

$v = 12 m . s^{- 1}$ $v=12 m.s^(-1)$

$a = g \approx 10 m . s^{- 1}$ $a=g~~10 m.s^(-1)$

$v = u + a t$ $v=u+at$

$\Rightarrow 12 = 6 + 10 t$ $rArr 12=6+10t$

$t = 0.6 s$ $t=0.6 s$

(b)
$u = 12 m . s^{- 1}$ $u=12 m.s^(-1)$

According to question,

$v = 24 m . s^{- 1}$ $v=24 m.s^(-1)$

$a = g \approx 10 m . s^{- 1}$ $a=g~~10 m.s^(-1)$

$v = u + a t$ $v=u+at$

$\Rightarrow 24 = 12 + 10 t$ $rArr 24=12+10t$

$t = 1.2 s$ $t=1.2 s$

Answer 2 · 2017-10-22T16:36:35

a) $t = 1.06 s$ $t = 1.06 s$
b) $t = 2.37 s$ $t = 2.37 s$

Explanation:

Speed is the magnitude of velocity. The fish will continue to have its horizontal speed at 6.0 m/s. Speed will be the sum of its increasing vertical speed and its constant horizontal speed.

Its increasing vertical velocity will be a*t downward. So its velocity, will be the vector sum of its vertical and horizontal velocities. Its speed can be found using Pythagoras
${(s p e e d)}^{2} = \sqrt{{(6.0 \frac{m}{s})}^{2} + {(a \cdot t)}^{2}}$ $(speed)^2 = sqrt((6.0 m/s)^2 + (a*t)^2$

a) We want the time at which the speed is 12 m/s, so

${(12 \frac{m}{s})}^{2} = \sqrt{{(6.0 \frac{m}{s})}^{2} + {(a \cdot t)}^{2}}$ $(12 m/s)^2 = sqrt((6.0 m/s)^2 + (a*t)^2$

$(144 - 36) \frac{m^{2}}{s^{2}} = {(a \cdot t)}^{2}$ $(144 - 36) m^2/s^2 = (a*t)^2$

Using 9.8 m/s^2 as the value of a,

$108 \frac{m^{2}}{s^{2}} = {(9.8 \frac{m}{s^{2}})}^{2} \cdot t^{2}$ $108 m^2/s^2 = (9.8 m/s^2)^2*t^2$

$t^{2} = ⎛ ⎝ \frac{108 \frac{m^{2}}{s^{2}}}{96.1 \frac{m^{2}}{s^{4}}} ⎞ ⎠ = 1.12 s^{2}$ $t^2 = ((108 m^2/s^2)/(96.1 m^2/s^4)) = 1.12 s^2$

$t = 1.06 s$ $t = 1.06 s$

b) We now want the time at which the speed is 24 m/s, so

${(576 \frac{m}{s})}^{2} = \sqrt{{(6.0 \frac{m}{s})}^{2} + {(a \cdot t)}^{2}}$ $(576 m/s)^2 = sqrt((6.0 m/s)^2 + (a*t)^2$

$(576 - 36) \frac{m^{2}}{s^{2}} = {(a \cdot t)}^{2}$ $(576 - 36) m^2/s^2 = (a*t)^2$

Using 9.8 m/s^2 as the value of a,

$540 \frac{m^{2}}{s^{2}} = {(9.8 \frac{m}{s^{2}})}^{2} \cdot t^{2}$ $540 m^2/s^2 = (9.8 m/s^2)^2*t^2$

$t^{2} = ⎛ ⎝ \frac{540 \frac{m^{2}}{s^{2}}}{96.1 \frac{m^{2}}{s^{4}}} ⎞ ⎠ = 5.62 s^{2}$ $t^2 = ((540 m^2/s^2)/(96.1 m^2/s^4)) = 5.62 s^2$

$t = 2.37 s$ $t = 2.37 s$

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Question #5ced9

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