1. An object moving with velocity 15 ms accelerates at 2 m/s for 5 seconds. Calculate the final velocity.
1 answer:
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Answer:
(a)
(b)
Explanation:
We can derive the initial speed of the rock from the equation of the speed in function of the time:
Using the given values for the speed at time t=1.7s, we get:
In words, the speed of the rock at launch is 34m/s (a).
Next, we use this to calculate the speed at t=4.9s:
This means that the speed of the rock at 4.9s after the launch is 14m/s (b), and the negative sign means that it is moving downwards.
Answer:
His pitching speed is 38 m/s.
Explanation:
Hi there!
Please see the attached figure for a better understanding of the problem.
The position of the ball at any time t is given by the following vector:
r = (x0 + v0 · t, y0 + 1/2 · g · t²)
Where:
r = position vector of the ball at time t.
x0 = initial horizontal position.
v0 = initial horizontal velocity.
t = time.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
Let's place the origin of the frame of reference at the throwing point so that x0 and y0 = 0.
When the ball reaches the ground, its position vector will be r1 (see figure). Using the equation of the vertical component of the position vector, we can find the time at which the ball reaches the ground. At that time, the horizontal component of the position is 30 m and the vertical component is -3.0 m (see figure):
y = y0 + 1/2 · g · t² (y0 = 0)
y = 1/2 · g · t²
-3.0 m = 1/2 · (-9.8 m/s²) · t²
-3.0 m / -4.9 m/s² = t²
t = 0.78 s
Now, knowing that at this time x = 30 m, we can find v0:
x = x0 + v0 · t (x0 = 0)
x = v0 · t
30 m = v0 · 0.78 s
v0 = 30 m / 0.78 s
v0 = 38 m/s
His pitching speed is 38 m/s.
