Velocity,Speed & Position of Spaceship After 5 Seconds

[amit25]

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

 

[wbandersonjr]

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.

 

[amit25]

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

 

[wbandersonjr]

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.

 

[asteriatic]

?

Based on the given information, we can calculate the velocity, speed, and position of the spaceship after 5 seconds. Since the spaceship is initially at rest, its initial velocity is 0 m/s. Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can calculate the velocity after 5 seconds as v = 0 + (1,2,3)*5 = (5,10,15) m/s. This means that the spaceship is moving at a velocity of 5 m/s along the x-axis, 10 m/s along the y-axis, and 15 m/s along the z-axis after 5 seconds.

The speed of the spaceship is the magnitude of its velocity, which can be calculated using the equation speed = |v| = √(v_x^2 + v_y^2 + v_z^2) = √(5^2 + 10^2 + 15^2) = √350 = 18.7 m/s.

To calculate the position of the spaceship after 5 seconds, we can use the equation s = ut + 1/2at^2, where s is the position, u is the initial velocity, a is the acceleration, and t is the time. Since the spaceship is initially at rest, its initial position is also 0. Therefore, the position after 5 seconds is s = 0 + 1/2*(1,2,3)*(5)^2 = (12.5, 25, 37.5) m. This means that the spaceship has traveled a distance of 12.5 m along the x-axis, 25 m along the y-axis, and 37.5 m along the z-axis in 5 seconds.

In conclusion, after 5 seconds, the velocity of the spaceship is (5,10,15) m/s, the speed is 18.7 m/s, and the position is (12.5, 25, 37.5) m. The total distance traveled by the spaceship in 5 seconds is 43.3 m.