Solve the problem involving the velocity - time graph

[chwala]

Homework Statement

see attached

Relevant Equations

Mechanics

For part (a)

I came up with a simultaneous equation, i.e

$m+x+4m+700$
$5m+x=700$

and

$15000=\dfrac{1}{2}[5m+2x]25$
$1200=5m+2x$

therefore on solving the simultaneous,

$5m+x=700$
$1200=5m+2x$

we get $x=500$ and $m=40$

the ms approach is here; more less similar approach.

Part (c) is straightforward.

For part (d) i used the concept on gradient,

i have for the deceleration part,

$m =\left[ \dfrac{25-0}{540-700} \right]$

$-0.15625=\left[\dfrac{y_1 - 0}{572 - 700}\right]$

$(-0.15625) (-128) =y_1$

$y_1 = 20$.

Mark scheme approach is here for part (d)

just sharing in case there is more insight.

 

[kuruman]

If you want people to comment on your solution, you should be more clear about what you are doing by (a) defining what the symbols you use stand for and (b) justifying the equations in which you use these symbols.

In other words explain what you are doing by placing yourself in the position of the reader.

 

[chwala]

If you want people to comment on your solution, you should be more clear about what you are doing by (a) defining what the symbols you use stand for and (b) justifying the equations in which you use these symbols.

In other words explain what you are doing by placing yourself in the position of the reader.

done.

 

[SammyS]

done.

Not completely.

Your solution to part (b) is well organized and we can infer what the variables represent by your labelling on your graph for part (a).
It would have been good for you to indicate that the area under the velocity-time graph gives the distance traveled, and that you were using units of metes for that, and that you computed the area using the formula for area of a trapezoid.
Also, your answers should definitely include units.

For part(d):
You have previously used the variable, $m$, for another purpose in parts (a) and (b). Besides, we have a name for this quantity, namely acceleration.

If you have made changes to the OP as a result of @kuruman 's comments, you should indicate that when you edit the OP. I know this has been pointed out on other occasions.

Homework Statement: see attached
Relevant Equations: Mechanics

For part (a)

View attachment 342548

I came up with a simultaneous equation, i.e

$m+x+4m+700$
$5m+x=700$

and

$15000=\dfrac{1}{2}[5m+2x]25$
$1200=5m+2x$

therefore on solving the simultaneous,

$5m+x=700$
$1200=5m+2x$

we get $x=500$ and $m=40$

the ms approach is here; more less similar approach.

View attachment 342546


Part (c) is straightforward.

For part (d) i used the concept on gradient,

i have for the deceleration part,

$m =\left[ \dfrac{25-0}{540-700} \right]$

$-0.15625=\left[\dfrac{y_1 - 0}{572 - 700}\right]$

$(-0.15625) (-128) =y_1$

$y_1 = 20$.

Mark scheme approach is here for part (d)

View attachment 342547

just sharing in case there is more insight.