x=x0+v0t+12at2
An object has a starting position of 2 m, a starting velocity of 15 m/s, and is moving at a constant speed. Which option shows the final position of the object after 1 s?
To find the final position of the object after 1 s, we can use the equation:
x = x0 + v0t + (1/2)at^2
Given:
x0 = 2 m (starting position)
v0 = 15 m/s (starting velocity)
t = 1 s (time)
We can substitute these values into the equation:
x = 2 + (15)(1) + (1/2)(0)(1)^2
x = 2 + 15 + 0
x = 17
Therefore, the final position of the object after 1 s is 17 m.
To find the final position of the object after 1 second, we can use the equation:
x = x0 + v0t + 1/2at^2
Given:
x0 = 2 m (starting position)
v0 = 15 m/s (starting velocity)
t = 1 s (time)
a = 0 (constant speed)
Plugging in the values, the equation becomes:
x = 2 + 15(1) + 1/2(0)(1)^2
Simplifying further:
x = 2 + 15 + 0
x = 17 m
Therefore, the final position of the object after 1 second is 17 meters.
The equation you provided is the equation of motion for an object in one dimension, where x is the final position, x0 is the initial position, v0 is the initial velocity, t is the time, and a is the acceleration.
To find the final position of the object after 1 second (t = 1s), you can plug the given values into the equation and solve for x.
The given values are:
x0 = 2 m (initial position)
v0 = 15 m/s (initial velocity)
t = 1 s (time)
a = 0 m/s^2 (since the object is moving at a constant speed, there is no acceleration)
Substituting these values into the equation, we get:
x = 2 + 15(1) + 0(1^2)
Simplifying:
x = 2 + 15 + 0
x = 17
Therefore, the final position of the object after 1 second is 17 meters.