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Murljashka [212]
3 years ago
15

8. What is the rate of change in an object’s position?

Physics
2 answers:
adelina 88 [10]3 years ago
4 0
A similar but separate notion is that of velocity, which the rate of change<span> of </span>position<span>. Example . If p(t) is the </span>position<span> of an </span>object<span> moving on a number line at time t (measured in minutes, say), then the average </span>rate of change<span> of p(t) is the average velocity of the </span>object<span>, measured in units per minute.</span>
Sloan [31]3 years ago
4 0
<span> the </span>rate<span> at which an object </span><span>changes position. but if you need a number or a formula i need to know if its on a hill. what panet. what the object is. etc. etc. etc.

</span>
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A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend.
MatroZZZ [7]

Answer: Part(a)=0.041 secs, Part(b)=0.041 secs

Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction

now we know that

a=-9.81 m/s^2  ( negative because it is pulling the player downwards)

we also know that

s=76 cm= 0.76 m ( maximum s)

using kinetic equation

v^2=u^2+2as

where v is final velocity which is zero at max height and u is it initial

hence

u^2=-2(-9.81)*0.76

u=3.8615 m/s\\

now we can find time in the 15 cm ascent

s=ut+0.5at^2

0.15=3.861*t+0.5*9.81t^2\\

using quadratic formula

t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}

t=0.0409 sec

the answer for the part b will be the same

To find the answer for the part b we can find the velocity at 15 cm height similarly using

v^2=u^2+2as

where s=0.76-0.15

as the player has traveled the above distance to reach 15cm to the bottom

v^2=0^2 +2*(9.81)*(0.76-0.15)

v=3.4595

when the player reaches the bottom it has the same velocity with which it started which is 3.861

hence the time required to reach the bottom 15cm is

t=\frac{3.861-3.4595}{9.81}

t=0.0409

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