Hello it's a me again drifter1! Today we continue with Physics and get into the promised exercises for Rectlinear motion. You should check the posts before to see the Theory behind all that stuff! So, without further do, let's get straight into it!
To keep it organized let's give the exercises names depending on the topic they are about!
Suppose an object passes 100m in 9.83 seconds.
At the segment of 50m to 70m the time interval was 1.7 seconds.
Which is the average velocity for
a) The whole distance
b) The segment of 50m to 70m
The average velocity is being calculated using u avg = Δx / Δt.
For the whole distance we have:
u avg = 100 / 9.83 = ~ 10.17 m/s
For the segment we have:
u avg = 70 - 50 / 1.7 = ~ 11.76 m/s
An object's position in time is given by the equation: x = 1.4*t^2 + 0.1*t^3 m
a) Calculate the average velocity at the time interval t = 0s to t= 10s
b) Calculate the instantaneous velocity at
i) t = 0s, ii) t = 5s, iii) t = 10s
If we differentiate the equation of position we get the equation for velocity that looks like this:
u = 2.8t + 0.3*t^2 m/s
The average velocity is also equal to u avg = u2 - u1 / 2.
This means that we will calculate the (instantaneous) velocity at t = 0s and t = 10s and then use this equation to find the average velocity.
u at 0s = 2.8*0 + 0.3*0^2 = 0 m/s
u at 10s = 2.8*10 + 0.3*10^2 = 28 + 30 = 58 m/s
So, u avg = 58 - 0 / 2 = 29 m/s
To find the instantaneous velocity we can simply use the same equation for each time.
We already did i) u at 0s = 0 m/s and iii) u at 10s = 58 m/s
ii) u at 5s = 2.8*5 + 0.3*5^2 = 21.5 m/s
Average and instantaneous acceleration:
The velocity of an object in time is given by: u(t) = 5 + 0.2*t^2 m/s
a) Find the average acceleration for the interval t = 0s to t = 5s
b) Find the instantaneous acceleration at t = 2.5 s
By differentiating the velocity we get the acceleration function:
a(t) = 0.4*t m/s^2
The average acceleration can be given by a avg = a2 - a1 /2.
So, we will calculate the instantaneous acceleration at t = 0s and t = 5s and use the equation.
a(0) = 0 m/s^2
a(5) = 0.4*5 = 2 m/s^2
So, a avg = 2 - 0 / 2 = 1 m/s^2
At t = 2.5 seconds the acceleration is:
a(2.5) = 0.4*2.5 = 1 m/s^2
We can see that the acceleration at the center of the time interval equals the average acceleration, but this is just an occurrence and doesn't have to happen every time and is not a theorem or anything like that!
Constant acceleration motion:
An object starts with zero velocity (immobile) and travels a distance of 480m in 16 seconds, having a constant acceleration the whole time period.
Calculate the velocity at t = 16s.
The acceleration is constant, but unknown which means that we will use one of the special equations!
x - x0 = 1/2 (u0 + u) t =>
480 - 0 = 1/2 (0 + u)*16 =>
16u/2 = 480 =>
u = 480 / 8 = 60 m/s
Object free fall:
An object gets to an max height of 0.64m when being thrown from the ground to the air.
a) How much is the velocity it gets thrown with?
b) How long does it stay in air?
Suppose the gravity acceleration is 9.8 m/s^2.
The acceleration is constant an known, but we don't know the time.
This means that we will use one of the special equations.
u^2 = u0^2 + 2(-g)(x - x0) =>
0 = u0^2 + 2*(-9.8)(0.64 - 0) =>
u0^2 = 2*9.8*0.64 =>
u0^2 = 12.544 => u0 = root(12.544) => u0 = ~ 3.54 m/s
By using the simple velocity equation we get:
u = u0 +(-g)t =>
0 = 3.54 -9.8t =>
t = 3.54/9.8 => t = ~ 0.36 sec
This is the time needed to get to that height.
When crashing down the object has a velocity of:
u^2 = u0^2 + 2(-g)(x - x0) =>
u^2 = 0 + 2*(-9.8)(0 - 0.64) =>
u^2 = 12.544 => u = 3.54 m/s down
So, the time going down is:
u = u0 + gt =>
3.54 = 0 + 9.8t =>
t = 0.36s, which is exactly the same as before, cause the time needed when going up is the same needed when going down, and the calculation is actually not needed to be done again!
So, the time in air is:
t in air = 0.36 *2 = 0.72 seconds
Velocity and position calculation using integration:
The acceleration is being given by the equation: a = 1.9t - 0.12t^2 m/s^2.
At t = 0 the object has zero velocity and is at x = 0m.
a) Find the functions of position and velocity
b) Calculate the maximum velocity the object reaches
Integrating the acceleration we get the velocity:
u = u0 + integral 0->t (1.9t - 0.12t^2)dt =>u = 1.9t^2/2 - 0.12t^3/3 m/s
Integrating the velocity we get the position:
x = x0 + integral 0->t (1.9t^2/2 - 0.12t^3/3)dt =>x = 1.9t^3/6 - 0.12t^4/12 m
The velocity is max at the point where the acceleration is 0.
1.9t - 0.12t^2 = 0 =>
t(1.9 - 0.12t) = 0 =>
t = 0s
1.9 - 0.12 t = 0 => t = 1.9/0.12 seconds
So, u max = 1.9*(1.9/0.12)^2/2 - 0.12*(1.9/0.12)^3/3 = ~ 79.39 m/s
And this is actually it for today and I hope that you enjoyed it!
From next time in Physics we will start getting into Plane motion that is in 2 dimensions instead of 1 and I will also start posting about other stuff in between of Physics!