The back wall of a home aquarium is a mirror that is a distance of 46.0 cm away from the front wall. The walls of the tank are n
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(a) The turtle's initial velocity is 4 m/s, initial position of the turtle is 5 cm, and initial acceleration is -1.25 m/s².
(b) The time when the velocity of the turtle is zero is 3.2 s.
(c) The time taken for the turtle to return to its starting point is 6.4 s.
(d) The time taken for the turtle to travel 30 cm is 0.08 s.
<h3>Initial velocity of the turtle</h3>The initial velocity of the turtle is calculated as follows;
The initial acceleration of the turtle is calculated as follows;
x(t) = 5 + 4t - 0.625t²
x(0) = 5 cm
<h3>Time when the velocity becomes zero</h3>v(t) = 4 - 1.25t
0 = 4 - 1.25t
1.25t = 4
t = 4/1.25
t = 3.2 s
<h3>Time taken to return to starting point</h3>The total distance traveled is calculated as follows
v² = u² + 2ad
0 = (4)² + 2(-1.25)d
0 = 16 - 2.5d
2.5d = 16
d = 16/2.5
d = 6.4 m
Time to travel the given distance;
d = ut + ¹/₂at²
6.4 = (4)t + ¹/₂(-1.25)t²
6.4 = 4t - 0.625t²
0.625t² - 4t + 6.4 = 0
solve the quadratic equation using formula method;
t = 3.2 s
The time travel the distance two times, = 2 x 3.2 s = 6.4 s
<h3>Time taken for the turtle to travel 30 cm</h3>d = ut + ¹/₂at²
0.3 = (4)t + ¹/₂(-1.25)t²
0.3 = 4t - 0.625t²
0.625t² - 4t + 0.3 = 0
solve the quadratic equation using formula method;
t = 0.08 s
Learn more about velocity here: brainly.com/question/6504879