Velocity,Speed & Position of Spaceship After 5 Seconds

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The spaceship, initially at rest, experiences a constant acceleration of (1, 2, 3) m/s². After 5 seconds, the velocity is calculated as (5, 10, 15) m/s using the equation v = at. The position after 5 seconds can be determined using the equation d = 1/2at², which provides the position vector. The discussion emphasizes the distinction between speed and distance as scalars, and velocity and position as vectors. Understanding these relationships is crucial for solving kinematics problems effectively.
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Spaceship initially at rest as measured,experiences constant acceleration of a=(1,2,3)m/s^2?

what is the velocity,after 5 seconds,
and speed after 5 seconds
and position after 5 secs
what is the distance it has traveled in 5secs
 
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What kinematics equations do you know that might apply to this problem? There is one that involves only velocity, acceleration and time which is useful for the first part.

Then there is also an equation involving position position, velocity, acceleration, and time which helps you solve the third part.

Finally, think about how speed and velocity are related; and how distance and position are related.
 
for the first one i used v=at
i got (5,10,15)m/s
the third part i used d=1/2at^2
im not sure how to find the distance traveled in 5 seconds
 
Ok, good you used the right equations and got the velocity and position vectors. Now, how is speed and distance related to velocity and position vectors? Think about whether speed and distance are vectors or scalars.
 
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