Let's say that a rock was thrown upward at time t = 0 with an initial velocity of 27 meters per second at a height of 50 meters. The only accln after being thrown is g which means a(t) = -9.8 m/sec^2 (If the upward direction is chosen to be positive.) v(0) = 27 m/sec, s(0) = 50 meters.
Integrate -9.8 wrt t to get a velocity function
v(t) = -9.8t + C where C is some constant. But we know that v(0) = 27, substitute:
v(0) = -9.8(0) + C = 27, so C = 27 and v(t) = -9.8t + 27
Integrate v(t) wrt t to get s(t)
s(t) = -4.9t^2 + 27t + C but we know that s(0) = 50, substitute
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Verified answer
v'(t) = ∫s''(t) dt
a(t) = ∫v'(t) dt
However:
a(t) ≠ ∫∫s''(t) dt
Would be helpful to have a concrete example.
Let's say that a rock was thrown upward at time t = 0 with an initial velocity of 27 meters per second at a height of 50 meters. The only accln after being thrown is g which means a(t) = -9.8 m/sec^2 (If the upward direction is chosen to be positive.) v(0) = 27 m/sec, s(0) = 50 meters.
Integrate -9.8 wrt t to get a velocity function
v(t) = -9.8t + C where C is some constant. But we know that v(0) = 27, substitute:
v(0) = -9.8(0) + C = 27, so C = 27 and v(t) = -9.8t + 27
Integrate v(t) wrt t to get s(t)
s(t) = -4.9t^2 + 27t + C but we know that s(0) = 50, substitute
s(0) = -4.9(0)^2 + 27(0) + C = 50
so C = 50 and s(t) = -4.9t^2 + 27t + 50
A simple example.
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