A rocket is fired vertically and reaches a speed of 34.3 m/s in half a second. One and a half seconds later the rocket’s speed is 19.6 m/s. Find the acceleration (in m/s2) and initial velocity (in m/s) of the rocket. Complete the equation of motion for the velocity of the rocket:
v(t) = t +
Assume a standard coordinate system where the upward vertical direction is positive and downward vertical direction is negative.
Record your answers to one decimal place.
To find the acceleration and initial velocity of the rocket, we can use the equations of motion for constant acceleration.
Let's denote the initial velocity of the rocket as "u" (in m/s), the final velocity as "v" (in m/s), the acceleration as "a" (in m/s^2), and the time interval as "t" (in seconds).
We are given:
Final velocity (v) at t = 0.5 seconds: v = 34.3 m/s
Final velocity (v) at t = 2 seconds: v = 19.6 m/s
The equation of motion that relates these variables is:
v = u + at
To find the acceleration (a), we can rearrange the equation as:
a = (v - u) / t
Substituting the values we know:
a = (19.6 m/s - 34.3 m/s) / (2 s - 0.5 s)
a = (-14.7 m/s) / (1.5 s)
a ≈ -9.8 m/s^2
So, the acceleration of the rocket is approximately -9.8 m/s^2 (negative sign indicates the direction of acceleration: downward).
To find the initial velocity (u), we can rearrange the equation as:
u = v - at
Using the values we know:
u = 34.3 m/s - (-9.8 m/s^2)(0.5 s)
u = 34.3 m/s + 4.9 m/s
u ≈ 39.2 m/s
So, the initial velocity of the rocket is approximately 39.2 m/s (positive sign indicates the direction of initial velocity: upward).
The equation of motion for the velocity of the rocket is:
v(t) = t + u
Substituting the value of u = 39.2 m/s, the equation becomes:
v(t) = t + 39.2
see question above, same procedure.
coincidence?
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