A runner slows down from a 9.50 m/s at a rate of 2.30 m/s^2 . (a) How far does she travel in the next 6.00 s? (b) What is her fi
1 answer:
Answer:
(a) d = 15.6 m
(b) v' = - 4.3 m/s
Given:
Initial velocity, v = 9.50 m/s
Acceleration, a = or deceleration = 2.30
Solution:
(a) For the calculation of the distance covered, we use eqn (2) of motion:
where
d = distance covered in time t
t = 6 s (Given)
Now,
d = 15.6 m
(b) For the calculation of her final velocity, we use eqn 1 of motion:
v' = v + at
v' = 9.50 + (- 2.30)(6) = - 4.3 m/s
(c) Since, the final velocity after the body slows down comes out to be negative and the distance covered and displacement of the body are different, it does not make sense.
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Answer:
Your friend has to wait 0.26 s after you throw the ball to start running.
Explanation:
The equation that gives the position vector of the ball is as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t ·sin α + 1/2 · g · t²)
Where:
x0 = initial horizontal positon
v0 = initial velocity
t = time
α = throwing angle
y0 = initial vertical position
g = acceleration due to gravity
The equation of displacement of your friend is as follows:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of your friend at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
Please, see the attached figure for a description of the situation. Notice that the frame of reference is located at the throwing point.
Let´s find the time of flight of the ball. We know that at the final time, the y-component of the vector r has to be -6.00 m (1 m above the ground). Then:
y = y0 + v0 · t ·sin α + 1/2 · g · t²
-6.00 m = 0 m + 9.00 m/s · t · sin 33.0° - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² · t² + 9.00 m/s · sin 33.0° · t + 6.00 m
Solving the quadratic equation:
t = 1.71 s
Now that we have the time of flight, we can calculate the x-component of the vector r (the horizontal distance traveled by the ball):
x= x0 + v0 · t · cos α
x = 0m + 9.00 m/s · 1.71 s · cos 33°
x = 12.9 m
Then, your friend will have to run (12.9 m - 11.0 m) 1.9 m to catch the ball 1 m above the ground.
Let´s see, how much time it takes your friend to run that distance:
x = x0 + v0 · t + 1/2 · a · t² (x0 = 0, v0 = 0)
x = 1/2 · a · t²
1.9 m = 1/2 · 1.80 m/s² · t²
Solving for t
t = 1.45 s
Then, since the time of flight of the ball is 1.71 s, your friend has to wait
1.71 s - 1.45 s = 0.26 s after you throw the ball to start running.
