Anti-Derivatives and Word Problems?

Question

This is the word problem:

On the moon, the acceleration due to gravity is 1.6 m/sec^2.

a) If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 30 sec. later?

b) How far below the point of release is the bottom of the crevasse?

c) If instead of being released from rest, the rock is thrown into the crevasse from the same point with a downward velocity of 4 m/sec, when will it hit the bottom and how fast will it be going?

THANK YOU SO MUCH! :]

Answer 1

a=dv/dt

dv/dt=1.6
dv=-1.6dt
Integrate
v=-1.6t+c
c is the initial velocity (0)

v=1.6(30)

v=dx/dt
dx/dt=-1.6t+c
dx=(-1.6t+c)dt
Integrate
x= -.8t^2+ct+c(2)

c is still the initial velocity and it's still zero. c(2) is the initial height. We know the time of flight when x=0 so just plug in 30 and solve for c(2)


For part c, just set the initial velocity as -4.

Answer 2

Hi,
a) If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 30 sec. later?

v=at…(Where v is velocity and t is time.)
v=1.6(30)
=48 m/s

b) How far below the point of release is the bottom of the crevasse?
s= Vot +(1/2)at²
s=(1/2)at² (Since Vo is apparently zero.)
=(1/2)(1.6)30²
=720 m
c) If instead of being released from rest, the rock is thrown into the crevasse from the same point with a downward velocity of 4 m/sec, when will it hit the bottom and how fast will it be going?
How long will it take to hit the bottom?
First, let’s find t from the following quadratic equation:
s=Vot + (1/2)at²
720=4t+(1/2)1.6t²
0.8t² +4t-720 = 0
From the quadratic formula, we will have:
t= [-b+-√(b²-4ac)]/(2a)
=[-4 +-√((4)²-4(.8)(-720)/(2*0.8)
= -4 +-√[16+2304)/(1.6)
=[-4+-√(2320)/1.6]
=[-4+-√(16*(145)]/1.6
=[-4+-4√(145)]/1.6
The minus part of the radical obviously will result in negative time, so plug the expression with the plus radical into your calculator and you’ll get
t= 27.60 s

How fast will it be going?
V=Vo + at
=4 +1.6(27.60)
=48.16 m/s

Answer 3

i'm no longer acquainted with US distance parameters like miles, ft and such yet i'm going to help with the different problem. first you're able to transform the speeds to m/s. because it is km/h, so which you're able to divide it through 3600 sec and multiply it to 1000m meaning you're able to divide your km/h numbers through three.6 to cause them to m/s we've: 25 km/h = 6.ninety 4 m/s 80 km/h = 22.22 m/s a) all of us comprehend V = a.t + V0 so -> 22.22 = a.13 + 6.ninety 4 => a = (22.22-6.ninety 4)/13 = a million.18 m/s^2 b) from the formulation x = a million/2.a.t^2 + V0.t + x0 -> we assume that the beginning element is 0 so x0 = 0 -> x = 0.5 * a million.18 * (13^2) + 6.ninety 4 * (13) + 0 = ninety 9.seventy one + ninety.22 + 0 = 189.ninety 3 m so our automobile starts with 25 km/h, hastens to 80 km/h through the acceleration unit of a million.18 m/s^2 and through that element is going almost one hundred ninety meters. P.S. be conscious that Rahul has calculated the two issues through assuming the numbers being meters and seconds and that's no longer nicely suited (solutions are incorrect) yet his technique is all nicely suited. convert and use them.