do you have a formulai can do this problem
204 ft = 62.2 m
We know that the initial velocity of the rocket was 0, we can find its acceleration using:
s = ut + 1/2 at²
62.2 = 0.5a(6)²
a = 3.46 m/s²
Now, the final velocity of the rocket:
v = u + at
v = 3.46 x 6
v = 20.76 m/s = 68.1 ft/s
Answer:
21.75 m
Explanation:
t = Time taken for the car to slow down = 0.75 s
u = Initial velocity = 50 m/s
v = Final velocity = 8 m/s
s = Displacement
a = Acceleration
Equation of motion

Acceleration is -56 m/s²

The distance covered in the 0.75 seconds is 21.75 m
Answer:
s₁ = 0.022 m
Explanation:
From the law of conservation of momentum:

where,
m₁ = mass of hockey player = 97 kg
m₂ = mass of puck = 0.15 kg
u₁ = u₂ = initial velocities of puck and player = 0 m/s
v₁ = velocity of player after collision = ?
v₂ = velocity of puck after hitting = 48 m/s
Therefore,

negative sign here shows the opposite direction.
Now, we calculate the time taken by puck to move 14.5 m:

Now, the distance covered by the player in this time will be:

<u>s₁ = 0.022 m</u>