5-kg box is released from rest on a frictionless incline has a uniform acceleration of 0.24 m/s2, what is the velocity of the box after 4 seconds?
Well, well, well, the box seems to be quite the daredevil, accelerating down that incline! Let's see how fast it goes after 4 seconds.
We can start by using the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity (which is zero because the box was at rest), a is the acceleration, and t is the time.
Plugging in the values we know, we get: v = 0 + (0.24 m/s²) * 4s.
Simplifying this, we find: v = 0.96 m/s.
So, after 4 seconds, the box will be moving at a speed of 0.96 m/s. It's really picking up the pace!
To find the velocity of the box after 4 seconds, we can use the equation of motion for uniformly accelerated motion:
v = u + at
where:
v is the final velocity,
u is the initial velocity (which is 0 in this case, since the box is released from rest),
a is the acceleration, and
t is the time.
Given:
u = 0 m/s (initial velocity)
a = 0.24 m/s^2 (acceleration)
t = 4 s (time)
Using the equation, we can calculate the final velocity:
v = 0 + (0.24 m/s^2)(4 s)
v = 0 + 0.96 m/s
v = 0.96 m/s
Therefore, the velocity of the box after 4 seconds is 0.96 m/s.
To find the velocity of the box after 4 seconds, we can use the following formula:
v = u + at
Where:
v = final velocity
u = initial velocity (since the box is released from rest, u = 0)
a = acceleration
t = time elapsed
From the given information, we know that the acceleration (a) is 0.24 m/s^2 and the time (t) is 4 seconds. Substituting these values into the formula, we get:
v = 0 + (0.24 m/s^2) * (4 s)
v = 0 + 0.96 m/s
Therefore, the velocity of the box after 4 seconds is 0.96 m/s.
v = a t = 0.24 * 4
The rest is just there to confuse you.