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erik [133]
3 years ago
6

In 5 meters, a person running at 0.8 m/s accelerates at 1.6 m/s2. How fast 16 points

Physics
1 answer:
frutty [35]3 years ago
5 0

They were going at a velocity 4.07m/s

<u>Explanation:</u>

Distance s =5 m

initial velocity u= 0.8 m/s

Acceleration a =1.6m/s2

We have to calculate the velocity with which they were going afterwards i.e final velocity.

Use the equation of motion

v^2=u^2+2as\\=0.8^2+2\times 1.6\times 5\\=16.64\\v=4.07 m/s

They were going with a velocity 4.07 m/s afterwards.

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Ymorist [56]

This question is incomplete the complete question is

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

Answer:

(a) Xs=0.459m

(b) t=0.984 s

(c) Vc=6.65 m/s

Explanation:

(a) To reach maximum distance

g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\  x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m

(b) For Time

To find t we must find t1 and t2

as

t=t1+t2

For T1

t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s

For T2

x_{l}=Vbt+(1/2)gt_{2}^{2}\\   as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s

For Total Time

t=t1+t2

t=0.306+0.6789

t=0.984s

(c) To find Vc

Vc=Vb+gt2

Vc=(0)+(9.8)(0.6789)

Vc=6.65 m/s

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