How long does it take the car to move 30 m from rest?

[chwala]

Homework Statement

See attached.

Relevant Equations

Mechanics

In my approach,

$T - R = ma$

$4.5m - 4m = ma$

and

$s = \dfrac{1}{2} at^2$

$a=0.5$

therefore,

$30 = 0.5 ×0.5 ×t^2$

$t^2 = \sqrt {120} = 10.95$ seconds $≈11$ seconds.

Your insight is welcome or better approach...

 

[Doc Al]

Looks good to me!

 

[Mark44]

In my approach,
$T - R = ma$
$4.5m - 4m = ma$

So $0.5m = ma \Rightarrow a = 0.5$
It was somewhat confusing to me to see the equation $s = \frac 1 2 at^2$ thrown in before the conclusion stating the value of a.

and
$s = \dfrac{1}{2} at^2$
$a=0.5$
therefore,
$30 = 0.5 ×0.5 ×t^2$
$t^2 = \sqrt {120} = 10.95$ seconds $≈11$ seconds.

Your insight is welcome or better approach...

 

[Doc Al]

@Mark44 makes a good point. It's a good idea to present your work logically so there can be no doubt about the steps you took. This is especially important if you must hand in your work to be graded!

(I knew what you meant, but still.)

 

[jbriggs444]

I found it disturbing to have $m$ presented as a pure number rather than as a mass. This means that the usual physics formula: $F=ma$ is not dimensionally correct.

Nonetheless, turning a blind eye to the inconsistent units, the result of the calculation is correct.

 

[chwala]

I found it disturbing to have $m$ presented as a pure number rather than as a mass. This means that the usual physics formula: $F=ma$ is not dimensionally correct.

Nonetheless, turning a blind eye to the inconsistent units, the result of the calculation is correct.

But question has clearly indicated $m$ as the mass. I do not understand what you mean by stating that it is given as a number. The net force on the toy car is in equilibrium and therefore the usual physics formula $F= ma$ is still dimensionally correct.

 

[PeroK]

But question has clearly indicated $m$ as the mass. I do not understand what you mean by stating that it is given as a number. The net force on the toy car is in equilibrium and therefore the usual physics formula $F= ma$ is still dimensionally correct.

In the question, $m$ as specified is a number. E.g. if the car has a mass of $2kg$, then $m =2$. $m$ itself does not have units.

Note that the other data, such as specifying a force of $4m \ N$ only makes sense dimensionally if $m$ is a number.

I thought it was an odd construction, but not particularly disturbing.

 

[heroslayer99]

Bro was disturbed by physics