A particle experiences a constant acceleration that is north at 100 m/s^2. At t=0, its velocity vector is 60 m/s east. At what time will the magnitude of the velocity be 100 m/s?
The eastward component will remain 60 m/s at it acquires a northward component.
Thinking of a 3,4,5 right triangle, you should be able to see that the velocity magnitude (the hypotenuse of the vector velocity triangle) will be 100 m/s when the northward velocity component is 80 m/s.
The particle will achieve that velocity
when 100 m/s^2*t = 80 m/s
which is t = 0.8 s.
To solve this problem, we can use vector addition to determine the magnitude of the velocity at a certain time.
Given:
Acceleration (a) = 100 m/s^2 (north)
Initial velocity (v0) = 60 m/s (east)
Target velocity magnitude (v) = 100 m/s
We need to find the time (t) when the magnitude of the velocity is 100 m/s.
Step 1: Decompose the initial velocity into its x and y components.
The initial velocity vector is 60 m/s east. Since the acceleration is in the north direction, the x-component of the velocity remains constant at all times.
The x-component (vx) remains 60 m/s.
Step 2: Determine the y-component of the acceleration (ay).
Since the acceleration is in the positive north direction, the y-component of the acceleration is positive.
ay = 100 m/s^2
Step 3: Use the kinematic equation to find the y-component of velocity (vy) at time t.
We can use the equation:
vy = v0y + ayt
Since the particle only has an initial velocity in the x-direction, the y-component of the initial velocity (v0y) is 0.
vy = 0 + (100 m/s^2)(t)
vy = 100t m/s (north)
Step 4: Determine the magnitude of the velocity (v) using the Pythagorean theorem.
The magnitude of the velocity (v) is given by:
v = √((vx)^2 + (vy)^2)
Substituting the values we found earlier:
v = √((60 m/s)^2 + (100t m/s)^2)
Step 5: Solve for time (t).
We need to find the time when the magnitude of the velocity is 100 m/s (v = 100 m/s). So we can write the equation as:
100 m/s = √((60 m/s)^2 + (100t m/s)^2)
Squaring both sides of the equation:
(100 m/s)^2 = ((60 m/s)^2 + (100t m/s)^2)
Simplifying the equation:
10000 m^2/s^2 = 3600 m^2/s^2 + 10000t^2 m^2/s^2
Subtracting 3600 m^2/s^2 from both sides:
6400 m^2/s^2 = 10000t^2 m^2/s^2
Dividing both sides by 10000 m^2/s^2:
0.64 = t^2
Taking the square root of both sides:
t = √0.64
Therefore, the time (t) when the magnitude of the velocity is 100 m/s is t = 0.8 seconds.
To find the time when the magnitude of the velocity is 100 m/s, we need to determine how the velocity changes with time.
Given that the particle experiences a constant acceleration that is north at 100 m/s^2, we can assume that the acceleration is in the y-direction. Since the velocity vector is initially 60 m/s east, we can also assume that the initial velocity is in the x-direction.
Let's break down the problem into x and y components.
In the x-direction, the acceleration is 0 m/s^2 because there is no change in velocity in that direction. Therefore, the velocity in the x-direction remains constant at 60 m/s.
In the y-direction, the acceleration is 100 m/s^2 north. Since the acceleration is constant, we can use the kinematic equation:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In the y-direction, using the given values:
v = ?
u = 0 m/s (since the particle starts from rest in the y-direction because its initial velocity is only in the x-direction)
a = 100 m/s^2 (acceleration in the y-direction)
t = ?
Now, we can rearrange the equation to solve for time:
v = u + at
t = (v - u) / a
Substituting the values:
t = (100 m/s - 0 m/s) / 100 m/s^2
Simplifying the equation:
t = 1 s
Therefore, it will take 1 second for the magnitude of the velocity to reach 100 m/s.