A material point, moving uniformly accelerated in a straight line in one direction, during time t increased its speed 3 times, having traveled a path of 20 m. Find t if the acceleration of the point is 5 m / s².
S = 20 meters – the distance covered by the material point in time t;
a = 5 m / s ^ 2 – point acceleration;
v1 = 3 * v0 – the point has increased its speed 3 times in time t.
It is required to determine the time of movement of the point t.
Let’s find the initial speed of the body:
(v1 – v0) / a = t;
(3 * v0 – v0) / a = t;
2 * v0 / a = t.
S = v0 * t + a * t ^ 2/2 = v0 * 2 * v0 / a + a * 4 * v0 ^ 2 / (2 * a ^ 2) =
= 2 * v0 ^ 2 / a + 2 * v0 ^ 2 / a = 4 * v0 ^ 2 / a, hence:
v0 = (a * S / 4) ^ 0.5 = (5 * 20/4) ^ 0.5 = (5 * 5) ^ 0.5 = 25 ^ 0.5 = 5 m / s.
Then the time of movement of the body is equal to:
t = 2 * v0 / a = 2 * 5/5 = 2 seconds.
Answer: the time of body movement is 2 seconds.
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