   Chapter 3.R, Problem 63E

Chapter
Section
Textbook Problem

# A canister is dropped from a helicopter 500 m above the ground. Its parachute does not open, but the canister has been designed to withstand an impact velocity of 100 m/s. Will it burst?

To determine

To Check:

Whether the canister will burst in spite of its design to withstand the impact velocity

Explanation

1) Concept:

If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is Fx+c where c is an arbitrary constant.

at=ddxvt

vt=ddxst

2) Given:

v = 100 m/s, s0=500 m

3) Calculation:

Acceleration is equal to the force of gravity on earth which is 9.8 m/s

That is a0=-9.8m/s because the canister is dropped above the ground.

The general antiderivative of at usingrules of antiderivative

vt=-9.8t+C

So we have v0=0

Substitute t=0 in v(t) we get

v0=-9.8(0)+C

0=0+C

⇒ C=0

Therefore, substitute C=0 in vt we get

vt=-9.8t

The general antiderivative of vt usingrules of antiderivative is

st=-9.8t22+D

Now we have s0=500

Substitute t=0 in st we get

s0=-9.8(0)22+D

500=0+D

D=500

Therefore, substitute D=500 in st we get

st=-9

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