Help please, Physics homework ch2?

[匿名]

2017-08-25 9:40 pm

The position of a particle moving along the x axis depends on the time according to the equation x = ct3 - bt7, where x is in meters and t in seconds. Let c and b have numerical values 2.5 m/s3 and 1.1 m/s7, respectively. From t = 0.0 s to t = 1.5 s, (a) what is the displacement of the particle? Find its velocity at times (b) 1.0 s, (c) 2.0 s, (d) 3.0 s, and (e) 4.0 s. Find its acceleration at (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s.

回答 (1)

不用客氣

2017-08-25 10:21 pm

x = 2.5t³ - 1.1t⁷
velocity, dx/dt = 7.5t² - 7.7t⁶
acceleration, d²x/dt² = 15t - 46.2t⁵

(a)
From t = 0.0 s to 1.5 s, displacement = [2.5(1.5)³ - 1.1(1.5)⁷] - [2.5(0.0)³ - 1.1(0.0)]⁷ = -10 m (2 sig. fig.)

(b)
At t = 1.0 s, velocity = 7.5(1.0)² - 7.7(1.0)⁶ = -0.20 m/s (2 sig. fig.)

(c)
At t = 2.0s, velocity = 7.5(2.0)² - 7.7(2.0)⁶ = -460 m/s (2 sig. fig.)

(d)
At t = 3.0s, velocity = 7.5(3.0)² - 7.7(3.0)⁶ = -5500 m/s (2 sig. fig.)

(e)
At t = 4.0s, velocity = 7.5(4.0)² - 7.7(4.0)⁶ = -31000 m/s (2 sig. fig.)

(f)
At t = 1.0 s, acceleration = 15(1.0) - 46.2(1.0)⁵ = -31 m/s² (2 sig. fig.)

(g)
At t = 2.0 s, acceleration = 15(2.0) - 46.2(2.0)⁵ = -1500 m/s² (2 sig. fig.)

(h)
At t = 3.0 s, acceleration = 15(3.0) - 46.2(3.0)⁵ = -11000 m/s² (2 sig. fig.)

(i)
At t = 4.0 s, acceleration = 15(4.0) - 46.2(4.0)⁵ = -47000 m/s² (2 sig. fig.)

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