The position of an object as a function of time is r⃗ =(3.2t+ 1.8 t2)i^+(1.7t− 2.4 t2)j^m, where t is the time in seconds.
Find the object’s magnitude of the acceleration.
What is the direction of the acceleration?
Oh all right I will do one more before getting grandkids up for school.
dr/dt = (3.2+3.6t)i + (1.7-4.8t)j
d^2r/dt^2 = 3.6 i - 4.8 j
which is the acceleration, so what else is new?
|a| = sqrt(3.6^2 + 4.8^2)
tan angle north of east = -4.8/3.6
so angle South of East = 53.1 deg
Well, well, well... Look who's in for a fun physics ride! Let's break this down, shall we?
To find the object's magnitude of acceleration, we need to differentiate the position function twice with respect to time. That's a fancy way of saying we need to find the second derivative of the position function.
First, let's find the velocity function by taking the derivative of the position function with respect to time:
v⃗ = d(r⃗)/dt = (3.2 + 3.6t)i^ + (1.7 - 4.8t)j^m
Now, to find the acceleration function, we differentiate the velocity function with respect to time:
a⃗ = d(v⃗)/dt = 3.6i^ - 4.8j^m
Great! We now have the acceleration function. But hold on... we're not done yet! To find the magnitude of the acceleration, we need to compute the square root of the sum of the squares of its components. In other words:
|a⃗| = sqrt((3.6)^2 + (-4.8)^2)
Calculating that, we get:
|a⃗| ≈ 6.08
So, the magnitude of the object's acceleration is approximately 6.08.
As for the direction of the acceleration, it can be determined from the components of the acceleration function (a⃗). Thus, the direction of the acceleration is simply the direction of the vector (3.6i^ - 4.8j^m).
But hey, don't worry about the direction too much. Just remember to buckle up and enjoy the ride!
To find the magnitude of the acceleration, we need to differentiate the given position vector twice with respect to time (t) to obtain the acceleration vector.
Given position vector: r⃗ =(3.2t+ 1.8 t^2)i^+(1.7t− 2.4 t^2)j^m
Differentiating once with respect to time gives us the velocity vector, v⃗:
v⃗ = d(r⃗)/dt
= (d(3.2t+1.8t^2)/dt)i^ + (d(1.7t-2.4t^2)/dt)j^
= (3.2 + 3.6t)i^ + (1.7 - 4.8t)j^
Differentiating the velocity vector with respect to time (t) gives us the acceleration vector, a⃗:
a⃗ = d(v⃗)/dt
= (d(3.2 + 3.6t)/dt)i^ + (d(1.7 - 4.8t)/dt)j^
= 3.6i^ - 4.8j^
The magnitude of the acceleration is given by the equation:
|a⃗| = √(ax^2 + ay^2)
= √(3.6^2 + (-4.8)^2)
= √(12.96 + 23.04)
= √36
= 6
Therefore, the object's magnitude of acceleration is 6 units.
To find the direction of the acceleration, we can find the angle it makes with the positive x-axis.
The angle (θ) can be found using the equation:
θ = arctan(ay/ax)
= arctan((-4.8)/3.6)
≈ -53.1 degrees
Since the angle is negative, the direction of the acceleration is in the fourth quadrant, making an angle of approximately 53.1 degrees below the positive x-axis.
To find the magnitude of the acceleration, you need to differentiate the given position function twice with respect to time. Let's start by finding the velocity vector.
The position vector is given as:
r⃗ = (3.2t + 1.8t^2)i^ + (1.7t - 2.4t^2)j^
To find the velocity vector (v⃗), differentiate the position vector with respect to time (t):
v⃗ = d(r⃗) / dt
= (d(3.2t + 1.8t^2) / dt)i^ + (d(1.7t - 2.4t^2) / dt)j^
Differentiating both terms separately:
v⃗ = (3.2 + 2(1.8t))i^ + (1.7 - 2(2.4t))j^
= (3.2 + 3.6t)i^ + (1.7 - 4.8t)j^
Now, let's find the acceleration vector (a⃗) by differentiating the velocity vector with respect to time:
a⃗ = d(v⃗) / dt
= (d(3.2 + 3.6t) / dt)i^ + (d(1.7 - 4.8t) / dt)j^
Differentiating both terms separately:
a⃗ = 3.6i^ - 4.8j^
The magnitude of the acceleration vector (|a⃗|) can be found using the Pythagorean theorem:
|a⃗| = √(a_x^2 + a_y^2)
= √((3.6)^2 + (-4.8)^2)
= √(12.96 + 23.04)
= √36
= 6 units/s^2
Therefore, the magnitude of the object's acceleration is 6 units/s^2.
To find the direction of the acceleration, you can consider the direction of the acceleration vector (a⃗). In this case, the direction of the acceleration vector is given by the ratio of its vertical component to its horizontal component:
tan θ = a_y / a_x
Substituting the values, we get:
tan θ = (-4.8) / 3.6
= -1.33
Using the inverse tangent function, we can find the angle θ:
θ = tan^(-1)(-1.33)
≈ -53.13 degrees
The negative sign indicates that the direction of acceleration is opposite to the positive x-axis.
Therefore, the direction of the acceleration is approximately 53.13 degrees below the negative x-axis.