An object is moving in the x-direction at 1.1m/s when it undergoes an acceleration a⃗ =0.56j^m/s2.
Find its velocity vector after 4.8s .
Express your answer as a comma-separated pair of components. Express your answer using two significant figures
since v = at, clearly it is
1.1i + 0.56*4.8j
Well, this object seems to be quite ambitious, trying to move in the x-direction and accelerate in the j-direction at the same time! I must say, it's quite the multitasker.
Now, let's see if we can figure out its velocity vector after 4.8 seconds. Since the object is only accelerating in the j-direction, we can ignore the acceleration in the x-direction.
Using the equation dv⃗ = a⃗ dt, where dv⃗ is the change in velocity vector and dt is the change in time, we can calculate the change in velocity in the j-direction.
dv⃗ = a⃗ dt
Δv⃗ = 0.56(j m/s^2)(4.8 s)
Multiplying those numbers together, we find that the change in velocity in the j-direction is approximately 2.7 j m/s.
Since there is no acceleration in the x-direction, the velocity in the x-direction stays constant. Therefore, the velocity vector after 4.8 seconds is (1.1 m/s, 2.7 m/s).
Remember to always keep an eye on your multitasking object! Who knows where it might go next?
To find the velocity vector after 4.8 seconds, we can use the equation:
Vf = Vi + a * t
where Vf is the final velocity vector, Vi is the initial velocity vector, a is the acceleration vector, and t is the time.
Given:
Vi = 1.1 m/s in the x-direction
a = 0.56 j m/s^2
t = 4.8 s
Plugging the values into the equation, we have:
Vf = (1.1 m/s) + (0.56 j m/s^2) * (4.8 s)
Vf = 1.1 m/s + 2.688 j m/s
Dividing the equation into x and y components, we get:
Vf_x = 1.1 m/s
Vf_y = 2.688 m/s
Therefore, the velocity vector after 4.8 seconds is (1.1, 2.688) m/s.
To find the velocity vector after 4.8 seconds, we can use the following kinematic equation:
v = u + at
Where:
v is the final velocity vector,
u is the initial velocity vector,
a is the acceleration vector,
t is the time.
Given:
Initial velocity u = 1.1 m/s in the x-direction,
Acceleration a = 0.56 j m/s^2 (where j is the unit vector in the y-direction),
Time t = 4.8 s.
Let's substitute the given values into the equation:
v = 1.1m/s + (0.56 j m/s^2)(4.8s)
Now, calculate each component of the velocity vector separately:
In the x-direction:
v_x = 1.1m/s + 0 = 1.1m/s
In the y-direction:
v_y = (0.56 j m/s^2)(4.8s) = 2.688 j m/s
Combining both components, the velocity vector v is:
v = 1.1 m/s (in the x-direction) + 2.688 m/s (in the y-direction)
So, the velocity vector after 4.8 seconds is approximately (1.1, 2.7) m/s, rounded to two significant figures.