A ball with an initial velocity of 2 m/s rolls for a period of 3 seconds. If the ball is uniformly accelerating at a rate of 3 m
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Given :
Displacement , y = 0.75 m .
Angular acceleration , .
Initial angular velocity , .
To Find :
The value of vertical velocity after time t = 0.25 s .
Solution :
By equation of circular motion is given by :
Putting all given values we get :
Now , vertical velocity is given by :
Therefore , the numerical value of the vertical velocity of the car at time t=0.25 s is 4.90 m/s .
Hence , this is the required solution .
Remember your kinematic equations for constant acceleration. One of the equations is 
, where 
= final position, 
= initial position, 
= initial velocity, t = time, and a = acceleration.
Your initial position is where you initially were before you braked. That means = 100m. You final position is where you ended up after t seconds passed, so = 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was = 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
Your acceleration is approximately
.
Your initial position is where you initially were before you braked. That means
Your acceleration is approximately