Author Topic: A few kinematics questions (Read 1735 times) Tweet Share

jasoN- · « **on:** August 03, 2010, 05:11:33 pm »

ATM ALL QUESTONS SOLVED, see below for solutions
[Thank you to those who answered or attempted]
I have a couple of questions which I'm stuck on, any help would be greatly appreciated. (Might add some more when i get stuck again)

[SOLVED]1) A particle moves in a line so that the velocity v m/s is given by $v=1+e^-^2^x$
a) Find the position, x m, in terms of time t seconds ( $t\geq0$ ), given that x=0 when t=0
b) Hence find a when $t=\log_{e} 5$

Answers: a) $\frac{1}{2}\log_{e} (2e^2^t-1)$
b) $\frac{-100}{2401}$

[SOLVED]2) A stationary police motorcycle is passed by a car travelling at 72km/h. The motorcycle starts in pursuit three seconds later. Moving with constant acceleration for a distance of 300 metres, it reaches a speed of 108km/h, which it maintains. Find the time it takes the motorcyclist to catch the car.

(Remember: 1m/s = 18/5 km/h)
I got 39 seconds, answer is 36 seconds

jasoN- · « **Reply #1 on:** August 03, 2010, 07:00:34 pm »

[SOLVED]Another question
The brakes are applied in a car travelling in a straight line. The acceleration a m/s^2 of the car is given by
$a=-0.4\sqrt{225-v^2}$ . If the initial velocity of the car was 12 m/s, find an expression for v the velocity of the car in terms of t, the time after the brakes were first applied.

Answer: $15\cos[\cos^-^1\frac{4}{5}+\frac{2t}{5}]$

apparently $sin(sin^{-1}(a)+b) = cos(cos^{-1}(a)-b)$

98.40_for_sure · « **Reply #2 on:** August 03, 2010, 08:05:51 pm »

1a)
$\frac{dx}{dt}=1+e^{-2x}=\frac{e^{2x}+1}{e^{2x}}$
$\frac{dt}{dx}=\frac{e^{2x}}{e^{2x+1}}$
$t=\int \frac{e^{2x}}{e^{2x}+1}dx$
Now let:
$u=e^{2x}+1$
$\frac{du}{dx}=2e^{2x}$
$t=\frac{1}{2}log_{e}\left | e^{2x}+1 \right |+c$
Sub in x=0, t=0 and find c
$t=\frac{1}{2}log_{e}\left | \frac{e^{2x}+1}{2} \right |$
$e^{2t}=\frac{e^{2x}+1}{2}$
$2e^{2t}-1=e^{2x}$
$x=\frac{1}{2}log_{e}(2e^{2t}-1)$

jasoN- · « **Reply #3 on:** August 03, 2010, 08:10:34 pm »

thanks, i didn't think of simplifying it to a positive exponential

98.40_for_sure · « **Reply #4 on:** August 03, 2010, 08:16:20 pm »

EDIT: bad method
Just double derive the answer from part a), you should get
$\frac{d^{2}x}{dy^{2}}=a=\frac{-4e^{2x}}{(2e^{2x}-1)^{2}}$
Now sub in $t=log_{e}5$
And you get $\frac{-100}{2401}$

jasoN- · « **Reply #5 on:** August 03, 2010, 08:33:15 pm »

thanks again, this algebra gets pretty tricky at times

98.40_for_sure · « **Reply #6 on:** August 03, 2010, 08:34:43 pm »

For question 2, i also got 39 seconds

EDIT: "the motorcycle takes off 3 seconds after"

Thus 39 - 3 = 36 lol staring you in the face

jasoN- · « **Reply #7 on:** August 03, 2010, 08:39:00 pm »

lol epic face palm! i was thinking 'this is one of the easiest kinematics questions, books gotta be wrong', thanks for the responses

98.40_for_sure · « **Reply #8 on:** August 03, 2010, 08:40:50 pm »

I don't have time to type out question 3, but just let a = dv/dt
Take reciprocal of both sides, and find t in terms of v, find c, then rearrange to get v in terms of t

Martoman · « **Reply #9 on:** August 03, 2010, 09:01:49 pm »

Your answer is wrong. It has to be multiplied by 15..... this is according to my maths :smitten:

link: http://img153.imageshack.us/img153/4899/23201855.jpg

98.40_for_sure · « **Reply #10 on:** August 03, 2010, 09:06:37 pm »

Quote from: Martoman on August 03, 2010, 09:01:49 pm

Your answer is wrong. It has to be multiplied by 15..... this is according to my maths :smitten:

link: http://img153.imageshack.us/img153/4899/23201855.jpg

Yeah i got same

jasoN- · « **Reply #11 on:** August 03, 2010, 09:08:38 pm »

Yeah you are right, I realised it before and edited it.

solution:
$a=-\frac{2\sqrt{225-v^2}}{5}$
$\frac{dv}{dt}=-\frac{2\sqrt{225-v^2}}{5}$
$\frac{dt}{dv}=-\frac{5}{2\sqrt{225-v^2}}$
$\frac{dt}{dv}=\frac{5}{2}(-\frac{1}{\sqrt{225-v^2}})$
$t=\frac{5}{2}cos^{-1}\frac{v}{15}+c$
$At t=0, v=12:$
$c=-\frac{5}{2}cos^{-1}\frac{4}{5}$
$t=\frac{5}{2}(cos^{-1}\frac{v}{15}-cos^{-1}\frac{4}{5})$
$\frac{2t}{5}=cos^{-1}\frac{v}{15}-cos^{-1}\frac{4}{5}$
$cos^{-1}\frac{v}{15}=cos^{-1}\frac{4}{5}+\frac{2t}{5}$
$\frac{v}{15}=cos(cos^{-1}\frac{4}{5}+\frac{2t}{5})$
$v=15cos(cos^{-1}\frac{4}{5}+\frac{2t}{5})$

Martoman · « **Reply #12 on:** August 03, 2010, 09:21:39 pm »

heh didn't need to type it out

My handwriting actually isn't that awful................. is it?

jasoN- · « **Reply #13 on:** August 03, 2010, 09:31:25 pm »

Nah, I can read your handwriting.
I already typed out the solution before u posted (or during), practicing my latex anyway
thanks for dedicating time to maths

Martoman · « **Reply #14 on:** August 03, 2010, 09:35:51 pm »

heh its maths

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