A cart driven by a propeller with a constant acceleration

Think about what is happening at the position [itex]x = 11.8\text{m}[/itex]. Is the cart accelerating? What is it's velocity?

 

Try using this equation: [itex]x=x_0 + v_0 t + \frac{1}{2}at^2[\itex]

You are correct that the velocity is 0 because we KNOW that the cart has to go the other way. But what is the acceleration at [itex]x = 11.8\text{m}[\itex]? Acceleration, like velocity, can be negative. A negative acceleration is colloquially called "deceleration".

 

I usually get students who are confused about this to draw the v-t graph for the motion.
displacement is the area under the graph and acceleration is the slope of the graph ... it helps them get their heads around the motion. This one will just be a triangle.

To use the kinematic equation method, you start by listing what you know:

d = 11.8m
u = 4.85m/s
v = 0m/s
a = ?

Now find the kinematic equation with each of these letters in it, and none of the others.
Note: the equation in post #5 does not do this.

 

What simon said. It's well worth memorising the standard SUVAT equations of motion for constant acceleration.

http://www.thestudentroom.co.uk/wik...Equations_of_Motion_for_Constant_Acceleration

Aside: The displacement is normally given the letter "s" rather than "d" but matters not.

 

#5

Quote

 

I wouldn't use it either.

That looks like this equation (one of the standard equations of motion)...

V²=U²+2as.

where

V= final velocity = 0
U= initial velocity
S= displacement

You appear to have made a trivial mistake when using it to calculate "a".

Rearrange it carefully and you get..

a = -U²/2s

= -0.997m/s²

eg not +0.997m/s². A +ve number implies the cart is accelerating (eg getting faster). A -ve number implies it's decelerating.

 

So I've heard ... I've never been able to :( ...so I got good at drawing v-t diagrams.

not everywhere in the World it isn't :) I'm used to "s" referring to an arc-length and S being a surface area while d for displacement gets cumbersome with calculus.

Indeed ... it is a good idea for students to get used to that too.

I would have worked it like this:
The v-t graph for this is a line starting at t=0, v=u and ending at t=T and v=0 (note: not given the final time - doesn't matter).
Notice that the line has a negative slope (sloping downward to the right) and the area is a right-angled triangle.

The slope of this graph is the acceleration: a=-u/T (slope = rise over run)
The area under the graph is the displacement: s=(1/2)Tu (area = half base times height)

2 equations and two unknowns - use the second to get an expression for T and substitute into the first:

T = 2s/u (since you know s and u, most people make the mistake of computing T now. It's normally fine at this level but risks rounding-off errors later.)

a = -u/(2s/u) = -u²/2s

... this is the kinematic equation that was used without the useless bits.
It is faster to memorize the kinematic equations. However, that way means you have to memorize five specific equations and this way you need only remember basic geometry and how to draw graphs - both general skills applicable outside kinematics.

 

I never even tried. Very long time ago, I memorized the obvious ## v = v_0 + at ## and the less obvious (it was given way before any elements of calculus were given) ## s = v_0t + \frac {at^2} 2 ## and got good at massaging those (plus the energy-related equations) into whatever shape I needed.

And yes, it was ## s ## for displacement :)

 

Those are the ones most students end up memorising. The first is off the slope of the v-t graph (though you may have got it via y=mx+c) ... you only need 2 equations and your second one has a lyrical, singsong, quality that makes it easy to memorise. The "obvious" second one is the area of the general v-t graph done by triangle + rectangle or as a trapesium:

Trapesium: s=(v+u)T/2

A lot of the people who have trouble with the equation approach respond to the graphical one.