Checking over test answers

[spiral vapor]

Group B Problems:

1. A bus is moving at a speed of 45 km/h and is 30 m away from a traffic light when it turns red. The bus stops 5 seconds later.

Has the bus crossed the traffic light?

If so, by how much?

2. A car with a mass of 1,000 kg begins braking uniformly and stops after covering a distance of 25 m.

Determine:
a) The braking acceleration.
b) The braking force.

3. A ball with a mass of 10 kg and a moment of inertia of 0.04 kg·m² has an angular momentum of 0.5 kg·m²/s with respect to its axis.

What is the radius of the ball?

The moment of inertia is given by .

[spiral vapor]

These are my answers for said problems, could anybody verify if theyre correct?

[brave girder]

1. by 32.5 metres

[spiral vapor]

how come

[brave girder]

convert speed in m/s

12.5 m/s

distance covered in 5 seconds

12.5*5

62.5

62.5-30

32.5

[untold topaz]

I approve 👍

[brave girder]

2nd question is incomplete ig. velocity not given

[spiral vapor]

that is exactly how it is given in the problem book my professor used

i've verified all of the solutions theyre correct

(the textbook shows problems + answers sheet at the end of the book)

t=5s for second problem

[brave girder]

wait rlly?

[rare token]

the bus is presumably slowing down, as it is at rest after 5 s

[brave girder]

we dont know whether the retardation is constant so i didnt presume anything at all

[rare token]

assuming a uniform deceleration,

$a = \frac{v-u}{t} = \frac{0 - 12.5}{5} = -2.5\ \text{m s}^{-2}$

[brave girder]

tho i was thinking of that

[topaz iris]

Jay

[rare token]

it can't be travelling at 45 km/h for the 5 s and then be instantaneously not moving

[brave girder]

but how do we know retardation is constant

[rare token]

we don't, the question is poorly expressed in that regard... but the light goes red, the bus continues at the same speed and then instantaneously stops after 5 s isn't a reasonable interpretation of the question, in my view.

[brave girder]

logicc

yea ur prob right my bad

[rare token]

\begin{align*} \text{Distance bus travels}\ = s &= ut + \frac{1}{2}at^2 \ &= 12.5 \times 5 + \frac{1}{2} \times -2.5 \times 5^2 \ &= \frac{25\times 5}{2} - \frac{2.5 \times 25}{2} \ &= \frac{25}{2}\big(5 - 2.5\big) \ \text{So, distance bus travels}\ &= \frac{125}{4}\ \text{m} = 31.25\ \text{m} \end{align*}

[topaz iris]

Jay

[rare token]

So, q 1 is yes, by 1.25 m

I agree with @brave girder that q 2 is incomplete.

If you take the deceleration as $a\ \text{m s}^{-2} < 0$

and that the vehicle slowed from $u\ \text{m s}^{-1}$ in time $t\ \text{s}$,

you can show that $u = -at\ \text{m s}^{-1}$

and that the distance traveled, $s = 25\ \text{m}$, is given by

$2s = -at^2 \implies -50 = at^2$...

so, you have two equations in three unknowns.

You need to know either $t$ or $u$ and use that $ut = 50$

to then be able to determine $a$.

[topaz iris]

Jay

[rare token]

If it slowed from 36 km/h = 10 m/s, it stopped in 5 s with a deceleration of -2 m/s^2...

If it stopped in 2.5 s, it was travelling at 20 m/s = 72 km/h and decelerated at -8 m/s^2...

You can choose any positive u or t and deduce the other and the deceleration rate.

[spiral vapor]

t is 5 seconds

the text got cut off

my apologies

[rare token]

then deceleration was at 2 m/s^2 and force was 2 000 N.

[spiral vapor]

yuh

i got a 95 on the test

[rare token]

Well done!