An object moves in a straight line with uniform acceleration, the initial velocity is 0.5m/s, and the displacement in the 7th s is 4m more than that in the 5th s Find: (1) the magnitude of the acceleration a of the uniformly accelerated motion of the object (2) The displacement of the object in the first 5S

An object moves in a straight line with uniform acceleration, the initial velocity is 0.5m/s, and the displacement in the 7th s is 4m more than that in the 5th s Find: (1) the magnitude of the acceleration a of the uniformly accelerated motion of the object (2) The displacement of the object in the first 5S

(1) Acceleration according to X7 − X5 = 2at2: a = X7 − X5
2T2＝4
two × 1＝2m/s2．
(2) The displacement of the object in the first 5S is: x = v0t + 1
2at2＝0.5 × 5+1
two × two × 25m=27.5m．
Answer: (1) the acceleration a of the uniformly accelerated motion of the object is 2m / S2;
(2) The displacement of the object in the first 5S is 27.5m

An object moves in a straight line with uniform acceleration, the initial velocity is 0.5m/s, and the displacement in the 7th s is 4m more than that in the 5th s Find: (1) the magnitude of the acceleration a of the uniformly accelerated motion of the object (2) The displacement of the object in the first 5S

(1) Acceleration according to X7 − X5 = 2at2: a = X7 − X5
2T2＝4
two × 1＝2m/s2．
(2) The displacement of the object in the first 5S is: x = v0t + 1
2at2＝0.5 × 5+1
two × two × 25m=27.5m．
Answer: (1) the acceleration a of the uniformly accelerated motion of the object is 2m / S2;
(2) The displacement of the object in the first 5S is 27.5m

An object moves in a straight line with uniform acceleration, the initial velocity is 0.5m/s, and the displacement in the 7th s is 4m more than that in the 5th S. calculate the acceleration of the object What is the displacement of the object in 5S? Why 27.5m instead of 25m?

Assuming the acceleration is a, the speed after 1s is 0.5 + A, and the speed after 4S is 0.5 + 3a. The starting speed in 5S is 0.5 + 4a, and the final speed is 0.5 + 5A. After running 1 / 2 * 1s * (0.5 + 4A + 0.5 + 5a), the starting speed in 7S is 0.5 + 6a, and the final speed is 0.5 + 7a. After running 1 / 2 * 1s * (0.5 + 6A + 0.5 + 7a), so 1 / 2 * 1s * (0.5 + 6A

Solving indefinite integral: ∫ (e ^ 3x + e ^ x) DX / (e ^ 4x-e ^ 2x + 1)

∫(e^3x+e^x)dx/(e^4x-e^2x+1)
=∫(e^x+e^-x)dx/(e^2x-1+e^-2x)
=∫d(e^x-e^(-x)/[(e^x-e^-x)^2+1]
=arctan(e^x-e^(-x))+C

Find indefinite integral exp (- x ^ 2 + x)

The original function is not an elementary function. Should this integral belong to the type of probability error function? You can check the table. ∫ e ^ (- x ² + x) dx= ∫ e^[- (x ² - x)] dx= ∫ e^[- (x ² - x + 1/4 - 1/4)] dx= ∫ e^[- (x ² - x + 1/4) + 1/4] dx= ∫ e^[...

For the definite integral interval (0, positive infinity) of exponential function, the integrand function is e ^ (- x2)^

That's no problem. You can turn it into a double integral. Let the original formula = t, then t ²= ∫（0,+∞）e^(-x ²) dx ∫（0,+∞）e^(-t ²) dt = ∫∫e^(-x ²- t ²) Dxdt can be obtained by using polar coordinates ²= ∫（0,+∞）d α ∫（0,π/2...