in a slap shot,a hockey player accelerates the puck from a velocity of 5.5m/s to 36m/s. if the shot take 3.16*10^-2, find the distance
a = (Vf-Vo)/t = (36-5.5)/0.0316 = 965.2 m/s^2.
d = Vo*t + 0.5a*t^2 =
5.5*0.0316 + 0.5*965.2*0.0316^2 = 0.656 m.
something very sophisticated
To find the distance covered by the puck during the slap shot, we can use the formula:
distance = average velocity × time
Average velocity can be calculated by adding the initial and final velocities and dividing by 2.
average velocity = (initial velocity + final velocity) / 2
Let's plug in the given values into the equations:
initial velocity (u) = 5.5 m/s
final velocity (v) = 36 m/s
time (t) = 3.16 × 10^-2 s
average velocity = (5.5 + 36) / 2 = 20.75 m/s
Now we can calculate the distance:
distance = average velocity × time
distance = 20.75 m/s × 3.16 × 10^-2 s
To multiply these values, we'll keep the numbers in scientific notation format:
distance = 2.07 × 10^1 × 3.16 × 10^-2
distance = 2.07 × 3.16 × 10^1 × 10^-2
distance = 6.5352 × 10^-1
So, the distance covered by the puck during the slap shot is approximately 0.65352 meters, or 65.352 centimeters.