Question
Use the equation for motion to answer the question.
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 second, we can plug the given values into the equation for motion:
x = x0 + v0t + 1/2at^2
Given:
x0 = 2 m (starting position)
v0 = 15 m/s (starting velocity)
t = 1 s (time)
Plugging in the values, we get:
x = 2 + 15(1) + 1/2(0)(1)^2
Simplifying:
x = 2 + 15 + 0
x = 17
Therefore, the final position of the object after 1 second is 17 meters.
To find the final position of the object after 1 s, we can use the equation for motion:
x = x0 + v0t + (1/2)at^2
Given:
Initial position (x0) = 2 m
Initial velocity (v0) = 15 m/s
Time (t) = 1 s
Acceleration (a) = unknown (but given that the object is moving at a constant speed, the acceleration would be 0 m/s^2)
We can plug in these values into the equation and solve for x:
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 would be 17 m.
To find the final position of the object after 1 second, we can use the equation for motion:
x = x0 + v0t + (1/2)at^2
Given:
Initial position (x0) = 2 m
Initial velocity (v0) = 15 m/s
Time (t) = 1 s
Substituting these values into the equation:
x = 2 + 15(1) + (1/2)a(1)^2
Since the object is moving at a constant speed, the acceleration (a) is 0. Therefore, the equation becomes:
x = 2 + 15 + 0
Simplifying:
x = 17
So, the final position of the object after 1 second is 17 meters.