Velocity and Acceleration Quick Check
4 of 54 of 5 Items
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?
(1 point)
Responses
30 m
30 m
60 m
60 m
17 m
17 m
32 m
To find the final position of the object after 1s, we can use the equation for motion given:
x = x0 + v0t + 1/2at^2
Given:
x0 = 2 m (starting position)
v0 = 15 m/s (starting velocity)
t = 1 s
Substituting these values into the equation:
x = 2 + 15(1) + 1/2(0)(1)^2
x = 2 + 15 + 0
x = 17 m
Therefore, the final position of the object after 1s is 17 m.
To find the final position of the object after 1 second, we can substitute the given values into the equation for motion:
x = x0 + v0t + (1/2)at^2
Given:
x0 (initial position) = 2 m
v0 (initial velocity) = 15 m/s
t (time) = 1 s
a (acceleration) = 0 (since the object is moving at a constant speed)
Substituting 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 second is 17 m.
To find the final position of the object after 1 second using the equation for motion, you need to substitute the given values into the equation and solve for x.
Given:
x0 = 2 m (starting position)
v0 = 15 m/s (starting velocity)
t = 1 s (time)
a = 0 m/s^2 (constant speed)
Using the equation x = x0 + v0t + 1/2at^2, we can substitute the values:
x = 2 + (15 * 1) + (1/2 * 0 * 1^2)
x = 2 + 15 + 0
x = 17 m
Therefore, the final position of the object after 1 second is 17 m.