Author Topic: Kinematics (Read 16747 times) Tweet Share

d0minicz · « **Reply #90 on:** July 19, 2009, 09:51:25 pm »

hey guys can soemone help me with b) of that question please
just tips/hints or anything x]

thanks !!!

d0minicz · « **Reply #91 on:** July 20, 2009, 06:27:29 pm »

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.

thanks....

dcc · « **Reply #92 on:** July 20, 2009, 06:51:44 pm »

Rearrange to find $v^2$ in terms of $a$ , then take the branch of the square root which fits the situation. (We have 'brakes' applied here, what sort of acceleration is this? What is the sign of this acceleration?)

EDIT: I didn't actually read the question, so there's a bit more to it than that!

d0minicz · « **Reply #93 on:** July 20, 2009, 07:24:38 pm »

okay so i get V in terms of a ; how do i continue ?

thx x]

dcc · « **Reply #94 on:** July 20, 2009, 07:28:23 pm »

Now that I've actually read the question, perhaps consider that $a = \frac{dv}{dt}$ then solving the differential equation (by inverting, I presume).

d0minicz · « **Reply #95 on:** July 20, 2009, 07:58:41 pm »

ah yeah tried that ... can ya please show me your working if u could lols or anyone
cheers man

dcc · « **Reply #96 on:** July 20, 2009, 09:15:25 pm »

Quote from: d0minicz on July 20, 2009, 06:27:29 pm

$a= -0.4 \sqrt {225 - v^2},\: v(0) = 12\: ms^{-1}$ .

$\begin{align*} a = \frac{dv}{dt} &= -0.4 \sqrt{225 - v^2}\\ \implies \frac{dt}{dv} &= -\frac{10}{4 \sqrt{225 - v^2}}\\ \implies t &= \int - \frac{10}{4 \sqrt{225 - v^2}}\:dv \end{align*}$

From there, it should be obvious (inverse trig).

ADDED DV FOR SUPER CLARITY AND PRECISION

d0minicz · « **Reply #97 on:** July 20, 2009, 10:41:55 pm »

ah overlookd someting

thanks..

Mao · « **Reply #98 on:** July 22, 2009, 04:39:58 pm »

dcc: dv

dcc · « **Reply #99 on:** July 22, 2009, 04:45:41 pm »

dv was quite clearly implied.

d0minicz · « **Reply #100 on:** August 19, 2009, 09:17:18 pm »

A body moving in a straight line with velocity given by v= 4x-1, where x is its position coordinate with respect to the origin. Its acceleration when x=1 is ...?

Need to see steps here... thx

hyperblade01 · « **Reply #101 on:** August 19, 2009, 09:23:27 pm »

$\frac{dv}{dx} = 4$

$a = v \frac{dv}{dx}$

$a = (4x-1)*4$

$a = 16x-4$

When x =1

$a= 16(1)-4$

$a = 12m/s^2$

kamil9876 · « **Reply #102 on:** August 19, 2009, 09:26:42 pm »

differentiate both sides with respect to $t$ :

$\frac{d}{dt}(v)=\frac{d}{dt}(4x-1)$

$\implies a=4v $

$a=4(4x-1)$
$a=12$

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