a = (delta v)/t = (v - vo)/t = (25 m/s - 15 m/s)/(4 s) = 2.5 m/s2
So, your average velocity = distance/time = (2 km)/(1/20 h + 1/30 h) = (2 km)/(1/12 h) = 24 km/h
- What starting speed is needed so that a ball thrown straight up will return 6 seconds later? The ball will go up for 3 seconds and come back down for 3 seconds, and its starting speed equals its landing speed. The problem is really asking, then "How fast will a ball be going after it falls freely from rest for 3 seconds?".
a = 10 m/s2, t = 3 s, v = ?
v = at = (10 m/s2)(3 s) = 30 m/s - How high will it go? Well, the ball will go up for 3 seconds and stop (for an instant its velocity = 0 m/s), so the ball's trip back down is free fall from rest. Therefore:
= 0.5(10 m/s2)(3 s)2 = 45 m Since the ball fell 45 meters, it must have been 45 meters high.
d = 0.75 m, a = 10 m/s2, t = ?
The hang time is twice the time to fall = 2(0.385 s) = 0.77 s
First, you need to realize that the salmon goes up some distance d (that we want to know) and then falls from rest that same distance, so we can use the equation to find this distance. In order to find the distance it falls, we first need to know the time it takes to fall. Since the salmon's starting speed equals the speed it has when it gets back to the water, the time of fall is the time it takes an object to reach a speed of 5 m/s when falling from rest. We can calculate this from v = at. Step 1: Find the time to reach a speed of 5 m/s when falling freely from rest.
v = 5 m/s, a = 10 m/s2, t = ?
Since v = at, t = v/a = (5 m/s)/(10 m/s2) = 0.5 s Step 2: Find the distance fallen from rest in 0.5 s
a = 10 m/s2, t = 0.5s, d = ?
= 0.5(10 m/s2)(0.5 s)2 = 1.25 m Alternatively,
If it is thrown at an angle, at the top of its path, its vertical velocity will be zero, however its horizontal velocity will be the same as its initial horizontal velocity minus whatever loss in speed as a result of air friction at that point. We won't know what that is without more information.