SRL Archive - Science

Can someone please help me answer these three questions?
It'd really help if you showed me step by step what to do :)

1)

How high will a ball travel (from the height it is thrown with a velocity of 15m/s[UP]?
I can't seem to get the correct answer as listed :/

2)

A car is travelling east at a constant velocity. Then it accelerates at 2 m/s^2[West] for 7 seconds and travels 0.072 km during this time. Find the initial velocity.

Can't seem to find the correct formula to use :/

3)

A train rounds a corner and the engineer spies a damsel in distress tied to the the tracks ahead so he immediately applies the brakes. How long would it take him to stop the train from a velocity of 50 km/h [fwd] at an acceleration of 2.0m/s^2 [backwarrd]? If the damsel is 0.05 km away when the brakes are applied, what is her fate?

Any help appreciated.

for 3) use vf^2 = vi^2 + 2ad and same for 1)

1) You know that vf must equal zero at the top of the "flight" of the ball and vi = 15m/s and a = -9.81m/s^2

3) Same thing basically accept a = -2

Note that i didn't double check the calculation but the method should be right :p

From: Sin

1)

How high will a ball travel (from the height it is thrown with a velocity of 15m/s[UP]?
I can't seem to get the correct answer as listed :/

Use the kinematics equations.
Taking downward as negative, a = -9.81 ms-2 (gravitational acceleration)
u (initial velocity) = 15
v (final velocity) = 0 (at the top where it turns direction)

Applying the formula
v = u + at
0 = 15 + -9.81t
t = 1.5290

Applying the next formula
s = ut + 1/2 at^2
s (displacement) = 15 (1.5290) + 1/2 (-9.81) (1.5290 ^ 2)
s = 11.468 m

From: Sin

2)

A car is travelling east at a constant velocity. Then it accelerates at 2 m/s^2[West] for 7 seconds and travels 0.072 km during this time. Find the initial velocity.

Can't seem to find the correct formula to use :/

s = ut + 1/2 at^2
0.072 x 1000 = 7u + 1/2 (2) (7^2)
u = 1.4694 ms-1

From: Sin

3)

A train rounds a corner and the engineer spies a damsel in distress tied to the the tracks ahead so he immediately applies the brakes. How long would it take him to stop the train from a velocity of 50 km/h [fwd] at an acceleration of 2.0m/s^2 [backwarrd]? If the damsel is 0.05 km away when the brakes are applied, what is her fate?

Any help appreciated.

v = u + at
0 = 50 * 1000 / 3600 + (-2)(t)
t = 6.9444 s


EDIT: oops forgot the 2nd part:
v^2 = u^2 + 2as
0 = (50 * 1000 / 3600)^2 + 2(-2) s
s = 48.225 m < 50m

The damsel will be 1.775 m from death :)



Just for future references (when you learn about momentum etc), these kinematics equations are only applicable when the acceleration is constant.

Dam riwu, you beat me to it :(
@ sin what did you get these questions in
Maths (Mechanics ) or Physics?

I've seen these before, I had them in our schools FST Class (Functions Statistics and Trigonometry), I was like finally a question I can answer! but :ninja:'d @riwu

From: Enslaved
Dam riwu, you beat me to it :(
@ sin what did you get these questions in
Maths (Mechanics ) or Physics?

Grade 11 physics

From: riwu
Note that i didn't double check the calculation but the method should be right :p


Use the kinematics equations.
Taking downward as negative, a = -9.81 ms-2 (gravitational acceleration)
u (initial velocity) = 15
v (final velocity) = 0 (at the top where it turns direction)

Applying the formula
v = u + at
0 = 15 + -9.81t
t = 1.5290

Applying the next formula
s = ut + 1/2 at^2
s (displacement) = 15 (1.5290) + 1/2 (-9.81) (1.5290 ^ 2)
s = 11.468 m


s = ut + 1/2 at^2
0.072 x 1000 = 7u + 1/2 (2) (7^2)
u = 1.4694 ms-1



v = u + at
0 = 50 * 1000 / 3600 + (-2)(t)
t = 6.9444 s


EDIT: oops forgot the 2nd part:
v^2 = u^2 + 2as
0 = (50 * 1000 / 3600)^2 + 2(-2) s
s = 48.225 m < 50m

The damsel will be 1.775 m from death :)



Just for future references (when you learn about momentum etc), these kinematics equations are only applicable when the acceleration is constant.

Thanks!
Here's another one:

A small airplane is travelling at a velocity relative to the air of 100km/h [E]. A wind from the north is blowing at 20kn/h. Determine the position of the airplane after 2.5h of this motion. Find the average velocity of the airplane during the 2.5h interval.

Are you doing HL or SL?

Bearing = 90 + angle

Displacement = Bearing + Resultant vector * time as no acceleration = 254.951

From: Enslaved
Bearing = 90 + angle

Displacement = Bearing + Resultant vector * time as no acceleration = 254.951

Beat me to it, got the same answer tho

Got it! How'd you get that image Enslaved?

Wolfram mathematica, i drew it then annotated, its a dam expensive bit of software :( - like £90 a year for updates and subscriptions - plus £2095 for the software license or ($3323.51 USD using today's exchange rate(1:1.59)

Okay, got another one:

In the first complete bounce of the golf ball, the initial velocity of the ball was 4.36m/s[81 degrees above the horizontal]. Find a)
how long it took for the ball to land
b)how high the ball bounced
c) whats its range is

If you could use v1 as initial velocity, v2 as final vel., a as acceleration, t as time, and d as distance, as well as give me an explanation about the horizontal thing, it'd be great :x
@Enslaved
@riwu

@sin Split it up into componets and do a parametric plot. Note, on phone so bear with post edits and no fancy graphics
Edit: since this is constant acceleration, you would use suvat
Where
S = displacement
U = initial velocity(v1) in this case
V = end velocity (v2) in this case
A = acceleration
T = time
Main equations:
V=U+A*T
S=U*T+0.5A*T^2
S=0.5T(U+V)
V^2=U^2 + 2*A*S

Here you are going to use the last one to find vertical displacement
So firstly the vertical component = sin(81)*4.3
Now to rearrange the formula, since we know the highest point will be when V =0,
-2AS=U^2
G=-9.8
So S = U^2/19.6
So S= (4.36sin(81))^2/19.8 = ~0.936586m -> 0.937m
Then for the parametric plot
X=4.36 cos(81) t
Y=4.36 sin(81) t - 4.9 t ^2
T = x / 4.36 cos (81),
Sub t into y and solve the quadratic equation for y=0, this should give you 2 values of x, one of them being 0 the other being dusplacement in x, divide that by 4.36 cos(81) and u would get T

As i said this is more or less how u do it, just work out the values.

I was sick during the week that we learned about components :/

Im not done yet… You woke me up… …

Edit, rough thing done, back to bed, ty for killing my sleep :-)

I apologize for any confusion with V maybe looking like U or something.
Vx = initial velocity in X direction
Vy = initial velocity in Y direction
X = location/displacement in X direction
Y = location/displacement in Y direction
ax = accleration in X direction (0 and not used)
ay = accleration in Y direction (-9.8 m/s^2)
Y1 = initial height (0)
X1 = Initial x location (0)

By components he just means the X and Y portions of the velocities. basically the sides of a right triangle that make the velocity vector. (Vx = 4.36 cos81 and Vy = 4.36 sin81)

also, imo, physics -> math section.

edit. I realize i drew a guy hitting the ball, but eh. I didnt read the question too well. bounce. hit. same thing.

on a side note. If youve got time, watch this. http://www.youtube.com/watch?v=IM630Z8lho8 along with alot of the other videos. I dont mean to confuse you or anything by posting this, but its one of things to realize that all of our equations we use in school are built around flat reference frames, but in reality, the world is not flat, so our equations work for everyday things because they are technically just approximations.

@Turpinator
Issues with your working
Never use the value of T to work out other values ( most of the time it is rounded) hence you get a diviation from the standard result), if you do, make sure that it is in the form of an equation and not a rounded value.
In Mine you would see that for the final process, I used a non-Rational value, this is Due to the Fact that Mathematica would take the value whole value that you pasted in (notice the ' at the end), as for the Sin and Cos Values, they were Rational and hence could be left as a decimal.

Due to this you can see that your Answer for max Y displacement is different

Mathematica

Edit: My method on solving range is through parametric plots. Makes life easier when there are other things to consider( no other things in this situation)