A small ball was projected horizontally at a velocity of 18.8 meters per second. It moved in a straight line with a uniform deceleration of 3.7 meters per second squared. Determine the velocity of the ball three seconds after it started moving.
In order to answer this question, we will use the equations of uniform acceleration known as the SUVAT equations. 𝑠 is the displacement of the particle, 𝑢, its initial velocity, 𝑣, the final velocity, 𝑎, the acceleration, and 𝑡 is the time. In this question, we are told that the initial velocity of the ball is 18.8 meters per second. The ball is decelerating at a rate of 3.7 meters per second squared. This means that 𝑎 is equal to negative 3.7.
We want to calculate the value of 𝑣 when the time is equal to three seconds. In order to do this, we will use the equation 𝑣 is equal to 𝑢 plus 𝑎𝑡. Substituting in our values, we have 𝑣 is equal to 18.8 plus negative 3.7 multiplied by three. Negative 3.7 multiplied by three is equal to negative 11.1. We need to subtract 11.1 from 18.8. This is equal to 7.7. The velocity of the ball three seconds after it started moving is 7.7 meters per second. We knew that this answer had to be less than the initial velocity of 18.8 as the ball is decelerating.