Final answer:
The problem asks for the displacement and velocity of a rock thrown downward from a bridge. By employing the kinematic equations for uniformly accelerated motion with the initial velocity and acceleration due to gravity, the displacement and velocity at various time intervals can be calculated.
Step-by-step explanation:
The question involves calculating the displacement and velocity at certain time intervals for a rock thrown down from a bridge. To solve this physics problem, we can use the kinematic equation for uniformly accelerated motion s = ut + (1/2)at2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. The initial velocity is given as 14.0 m/s (downward), and the acceleration due to gravity (which is downward) is 9.81 m/s2.
Let's calculate the displacement and velocity for each of the given times:
- (a) For t = 0.500 s: Displacement (s) is s = (14.0 m/s)(0.500 s) + (1/2)(9.81 m/s2)(0.500 s)2. Velocity (v) is v = u + at = 14.0 m/s + (9.81 m/s2)(0.500 s).
- (b) For t = 1.00 s: Displacement and velocity are calculated similarly, substituting t with 1.00 s.
- (c) For t = 1.50 s: Substitute t with 1.50 s in the kinematic formulas.
- (d) For t = 2.00 s: Use t = 2.00 s for the calculations.
- (e) For t = 2.50 s: Finally, calculate using t = 2.50 s.
Carry out these calculations to find the specific numerical values for displacement and velocity at each of the given times.