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A small piece of Styrofoam packing material is dropped from a height of 2.00 m above the ground. Until it reaches terminal speed, the magnitude of its accelera- tion is given by a g bv. After falling 0.500 m, the Styrofoam effectively reaches its terminal speed, and then takes 5.00 s more to reach the ground. (a) What is the value of the constant B? (b) What is the acceleration at t = 0? (c) What is the acceleration when the speed is 0.150 m/s?

Answer :

Answer:

Explanation:

Given

total height h=2 m

acceleration is given by

a=g-bv

after falling 0.5 m Styrofoam acquires its terminal velocity

remaining distance to fall is 2-0.5=1.5 m

time taken after acquiring terminal velocity is 5 s

[tex]v_{terminal}=\frac{1.5}{5}=0.3 m/s[/tex]

at terminal velocity a=0

thus g-bv=0

[tex]b=\frac{9.8}{0.3}=32.66[/tex]

at t=0 acceleration

[tex]a=9.8-32.66\cdot 0[/tex]

[tex]a=9.8 m/s^2[/tex]

(c)acceleration when speed is 0.15 m/s

[tex]a=9.8-32.66\cdot 0.15=4.89 m/s^2[/tex]

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