Answer:
a. Acceleration, a = 1.88 m/s²
b. Time, t = 7.87 seconds.
Explanation:
Given the following data;
Initial velocity, U = 14.5m/s
Final velocity, V = 29.3m/s
Distance, S = 172m
a. To find the acceleration of the speedboat;
We would use the third equation of motion;
V² = U² + 2aS
Substituting into the formula
29.3² = 14.5² + 2a*172
858.49 = 210.25 + 344a
344a = 858.49 - 210.25
344a = 648.24
a = 648.24/344
Acceleration, a = 1.88 m/s²
b. To find the time;
We would use the first equation of motion;
V = U + at
29.3 = 14.5 + 1.88t
1.88t = 29.3 - 14.5
1.88t = 14.8
Time, t = 14.8/1.88
Time, t = 7.87 seconds.
Answer:
Accuracy measures how close results are to the true or known value. Precision, on the other hand, measures how close results are to one another.
Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) = 
and
h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that
So on
= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
