caculating time

[COCoNuT]

A Honda Civic travels in a straight line along a road. Its distance xfrom a stop sign is given as a function of time t by the equation x(t) = 1.53m/s^2*t^2 - 5.10×10-2 m/s^3*t^3


Question #1. Calculate the average velocity of the car for the time interval t=0 to t1=1.93s. answer is in m/s

here's what i think i need to do:

plug in 0 and 1.93 for t... ok so if i plug in 0 for t, it would be zero, so i skip that and go on to plug in t1=1.93s for the equation.

x(t)= 1.53m/s^2(1.93)^2-5.10 X10^-2m/s^3(1.93s)^3
5.69m/s^2 - 36.66 X 10^-2 m/s^3
x(t)=5.3m/s^2 <--- can m/s^2 be substracted from m/s^3?

ok so would that be the time? how would i caculate the average velocity as stated in question #1?


i think i need to use the formula Vf(Final velocity) - V i(initial velocity)/time

ok so x(t) would be the time right? how would i find the final velocity and the initial velocity

[Tide]

To find the average speed you need to know the speed of the vehicle. Hint: differentiate!

[COCoNuT]

To find the average speed you need to know the speed of the vehicle. Hint: differentiate!


x(t) = 1.53m/s^2*t^2 - 5.10×10-2 m/s^3*t^3
ok so d/dt*x(t) =...

v(t) = 1.53(2)m/s*2(t) - 5.10 X 10^-2 (3)m/s^2*(3)t^2

is that correct? what am i suppose to do with v(t)?

[Tide]

Now find the average! Hint: Add up the values of the speed (aka integration!) and divide by the time interval!

[COCoNuT]

Now find the average! Hint: Add up the values of the speed (aka integration!) and divide by the time interval!

the time interval is x(t)=5.3m/s^2 right? did i even do that correctly?

and for the values of the speed, what am i integrating extactly? it cant be v(t), cause that would get me x(t), right? i am so confused

[needhelpperson]

A Honda Civic travels in a straight line along a road. Its distance xfrom a stop sign is given as a function of time t by the equation x(t) = 1.53m/s^2*t^2 - 5.10×10-2 m/s^3*t^3


Question #1. Calculate the average velocity of the car for the time interval t=0 to t1=1.93s. answer is in m/s

here's what i think i need to do:

plug in 0 and 1.93 for t... ok so if i plug in 0 for t, it would be zero, so i skip that and go on to plug in t1=1.93s for the equation.

x(t)= 1.53m/s^2(1.93)^2-5.10 X10^-2m/s^3(1.93s)^3
5.69m/s^2 - 36.66 X 10^-2 m/s^3
x(t)=5.3m/s^2 <--- can m/s^2 be substracted from m/s^3?

ok so would that be the time? how would i caculate the average velocity as stated in question #1?


i think i need to use the formula Vf(Final velocity) - V i(initial velocity)/time

ok so x(t) would be the time right? how would i find the final velocity and the initial velocity

first of all average velocity equals d/t so once you figure out
x(1.93) then divide it by 1.93 to get the average velocity.

you made some errors in your calc:

x(1.93)= 1.53m/s^2(1.93s)^2-5.10 X10^-2m/s^3(1.93s)^3

the units seconds cancel out, so you're left with meters.
5.69m - 36.66 X 10^-2 m

x = 5.3m

This is an easy question and doesn't require ne calculus, which i haven't taken yet.

[Tide]

That will work too but I guessed this was for a beginning calculus course. A more advanced student will recognize that \int \frac {dx}{dt} dt is just [itex]\Delta x[/tex]. I guess you're a more advanced student! ;-)

[COCoNuT]

first of all average velocity equals d/t so once you figure out
x(1.93) then divide it by 1.93 to get the average velocity.

you made some errors in your calc:

x(1.93)= 1.53m/s^2(1.93s)^2-5.10 X10^-2m/s^3(1.93s)^3

the units seconds cancel out, so you're left with meters.
5.69m - 36.66 X 10^-2 m

x = 5.3m

This is an easy question and doesn't require ne calculus, which i haven't taken yet.

ah perfect... thanks alot, it was easily than i thought. but how did the s^2 and s^3 cancel out?

[needhelpperson]

That will work too but I guessed this was for a beginning calculus course. A more advanced student will recognize that \int \frac {dx}{dt} dt is just [itex]\Delta x[/tex]. I guess you're a more advanced student! ;-)


Actually, I haven't done ne thing with calculus yet. I'm actually still in high school and i'm planning to take calculus this coming school year.

[needhelpperson]

ah perfect... thanks alot, it was easily than i thought. but how did the s^2 and s^3 cancel out?


x(1.93)= 1.53m/s^2(1.93s)^2-5.10 X10^-2m/s^3(1.93s)^3


As you can see, in (1.53m*(1.93s)^2)/s^2 and (10^-2*(1.93s)^3)/s^3

the units s^2/s^2 and s^3/s^3 = 1 so they can be ignored.

[Tide]

Actually, I haven't done ne thing with calculus yet. I'm actually still in high school and i'm planning to take calculus this coming school year.

Oops! Sorry about that. I KNOW you'll enjoy calculus! Keep up the good work.

[Chronos]

try A = \frac{1}{2gt^2}Therefore t^2=\frac{1}{2gA}

[Leong]

x(0)=0 m
x(1.93)=5.33 m
Average velocity, \bar{v}=\frac{x_{final}-x_{initial}}{t_{final}-t_{initial}}
=\frac{5.33-0}{1.93-0}

=2.76 m/s

[Chronos]

try A = \frac{1}{2gt^2}Therefore t^2=\frac{1}{2gA}I am latex challenged and used the wrong expression to boot. :blushing:
D = v_it+\frac{at^2}{2}

D = distance, m
v_i = initial velocity, m/s
t = time, s
a = acceleration, m/s

[robphy]

Now find the average! Hint: Add up the values of the speed (aka integration!) and divide by the time interval!


the time interval is x(t)=5.3m/s^2 right? did i even do that correctly?

and for the values of the speed, what am i integrating extactly? it cant be v(t), cause that would get me x(t), right? i am so confused

{\rm Average\ velocity}\ v_{avg}=\frac{\int v\ dt}{\int dt} = \frac{\Delta x}{\Delta t}
(This has units of velocity, as expected.)