HELP: A few question related to Applications of Calculus to the Physical World. (1 Viewer)

[Jackets]

Hi,

This is my first time posting a maths problem, and I'd be very greatful if you guys could help me with these questions!

1) The acceleration of a particle at any time, t seconds, is given by

a = -24t

a) If the particle starts from the origin with velocity 48 m/s, find expressions for velocity and displacement. Answer: v = -12t² + 48; x = -4t³ + 48t

b) When is the particle at rest? Answer: t = 2

c) Calculate the distance traveled between t = 4 and t = 7. Answer: 972 m.


3) A radioactive substance is decaying at a rate proportional to its mass. Its mass initially is measured at 37.4 grams. Twelve hours later the mass is measured at 31.7 grams.

Find: a) the decay constant; Answer: -0.014
b) the half life of the substance approx. 50 hours

4) A stone is dropped from a building 50 metres high. Its acceleration at any time, t, can be approximated by the equation a = -10.

a) Placing the origin at the foot of the building, derive equations for velocity and displacement. V = -10t; x = -5t² + 50

b) How long does it take for the stone to hit the ground at the foot of the building? What is its velocity then? 3.16 secs; -31.6 m/s

















Thanks!
​

[/LEFT]
[/CENTER]

 

[Azreil]

1a) a = -24t
v = Integral (-24t) dt
= -12t^2 + c
When t = 0, v = 48
48 = 0 + c
v = -12t^2 + 48
x = Integral (-12t^2 + 48)dt
= -4t^3 + 48t + c
When t = 0, x = 0
x = -4t^3 + 48t

b) At rest when v = 0
0 = -12t^2 + 48
12t^2 = 48
t^2 = 4
t = +2, -2
But t >= 0 therefore at rest when t = 2

c) Does not change direction as is only at rest when t = 2. (Not between 4 and 7).

x = -4t^3 + 48t ; when t=4
= -4(4)^3 + 48(4)
= -64
x = -4t^3 + 48t ; when t=7
= -4(7)^3 + 48(7)
= -1036

-1036 - -64 = 972m

2) Parabolic shape (roughly), minimum tp when resistance occurs.
dr/dt = negative until resistance occurs. Positive after resistance occurs.
d^2R/dt^2 = positive; concave up shape.

3)
Let t be time in hours, k be decay constant, M be mass and A be initial mass.
M=Ae^kt
a) Find k
31.7 = 37.4 * e^12k
e^12k = 0.847593582887700534759358 (woooo calculator)
ln e^12k = ln 0.847593582887700534759358
12k = ln 0.847593582887700534759358
k = ln 0.847593582887700534759358 / 12
= approx -0.0137795
b) 18.7 = 37.4 * e^kt
e^kt = 0.5
-0.0137795t = ln 0.5
t = ln0.5/-0.0137795
= 50.30
Therefore half life is approx 5 hours 18 minutes.

4) A stone is dropped from a building 50 metres high. Its acceleration at any time, t, can be approximated by the equation a = -10.

a) Placing the origin at the foot of the building, derive equations for velocity and displacement.

b) How long does it take for the stone to hit the ground at the foot of the building? What is its velocity then?

a) d^2y/dt^2 = -10
dy/dt = -10t + c
When t = 0, c = 0
dy/dt = -10t
y = -5t^2 + c
when t = 0, y = 50
y = -5t^2 + 50
b) hits ground when y = 0
0 = -5t^2 + 50
t^2 - 10 = 0
(t + sqrt10)(t - sqrt10)= 0
t = +sqrt10, - sqrt10
But t>=0 therefore t= +sqrt10
(if approximation -- t = 3.16)
dy/dt = -10 * sqrt10
(if approximation -- v = 31.6).

 

[Jackets]

Thank you so much Azreil! Anybody wanna answer the remaining ones?

 

[Azreil]

Was editing as I went, should all be up there now ;]

 

[Jackets]

Was editing as I went, should all be up there now ;]

Azreil, Thank you SO MUCH! You're a life saver!