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3 years ago

When a object is at rest, what is it’s speed??

Physics

2 answers:

Firdavs [7]3 years ago

7 0

Answer:

the speed when object at rest is 0 m/s

nadya68 [22]3 years ago

3 0

Answer:

0 (there is no speed)

Explanation:

If an object is at rest, it is not moving, and it doesn't have a speed, so the speed is zero.

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Read 2 more answers

A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^

xeze [42]

Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

$v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}$

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

$v_{100}=33.81$

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

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3 years ago

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.

SVEN [57.7K]

Answer:

a) Knowns, initial speed $v_{i}=13.0 m/s$ , final speed $v_{f}=0 m/s$ and gravity due it is a constant $g=9.8m/s^{2}$

b) The maximum high reached by the dolphin is $y_{max}=8.62 m$

c) Total time is $t=2.65s$

Explanation:

a) First of all the initial speed is given at the start of the problem, gravity is constant and final speed is known any object thrown straight up reaches its max high at $0m/s$ speed.

b) Second, now that we know final speed we use $v_{f} =v_{i}-gt$ , as we clear for $t=\frac{v_{i}-v_{f} }{g}=\frac{20.0m/s}{9.8m/s^{2} }=1.32 s$ .

Then we use $y=v_{i}t-\frac{1}{2} gt^{2}=(20.0m/s)(1.32s)-\frac{1}{2} (9.8m/s^{2} ) (1.32s)^{2} =8.62m$

c)Third, finally we can use $y=v_{i}t-\frac{1}{2} gt^{2}$ , as we know $y=0m$ when the dolphin fall into the water again and $v_{i} =13.0m/s$ , then we have $0=(13m/s)t-\frac{1}{2} (9.8m/s^{2} )t^{2}$ is a quadratic form $0=t(13.0-4.9t)$ so we have $t_{i}=0s$ and $t_{f}=\frac{13}{4.90} =2.65s$