Anonymous
May 14
A particle is moving in a straight line with acceleration A=sin2t+t^2 3 m/s^2. At time t=0,its velocity is 3m/s. The distance traveled by particle between t=0& t=π is?
A particle is moving in a straight line with acceleration A=sin2t+t^2 3 m/s^2. At time t=0,its velocity is 3m/s. The distance traveled by particle between t=0& t=π is?
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ChatGPT
May 14
To find the distance traveled by the particle between t=0 and t=π, we need to first find the velocity function and then integrate it to find the displacement function.
Given acceleration A = sin(2t) + t^2, we can find the velocity function by integrating the acceleration function with respect to time:
v(t) = ∫(sin(2t) + t^2) dt
v(t) = -1/2 cos(2t) + 1/3 t^3 + C
Given that the velocity at t=0 is 3 m/s, we can find the constant C:
v(0) = -1/2 cos(0) + 1/3 (0)^3 + C
3 = -1/2 + C
C = 3.5
Therefore, the velocity function is:
v(t) = -1/2 cos(2t) + 1/3 t^3 + 3.5
To find the distance traveled by the particle between t=0 and t=π, we need to integrate the velocity function over the interval [0, π]:
s(π) - s(0) = ∫[0, π] v(t) dt
s(π) - s(0) = ∫[0, π] (-1/2 cos(2t) + 1/3 t^3 + 3.5) dt
s(π) - s(0) = [-1/4 sin(2t) + 1/12 t^4 + 3.5t] [0, π]
s(π) - s(0) = [-1/4 sin(2π) + 1/12 π^4 + 3.5π] - [0]
Since sin(2π) = 0, the distance traveled by the particle between t=0 and t=π is:
s(π) = 1/12 π^4 + 3.5π
Therefore, the distance traveled by the particle between t=0 and t=π is 1/12 π^4 + 3.5π meters.
Given acceleration A = sin(2t) + t^2, we can find the velocity function by integrating the acceleration function with respect to time:
v(t) = ∫(sin(2t) + t^2) dt
v(t) = -1/2 cos(2t) + 1/3 t^3 + C
Given that the velocity at t=0 is 3 m/s, we can find the constant C:
v(0) = -1/2 cos(0) + 1/3 (0)^3 + C
3 = -1/2 + C
C = 3.5
Therefore, the velocity function is:
v(t) = -1/2 cos(2t) + 1/3 t^3 + 3.5
To find the distance traveled by the particle between t=0 and t=π, we need to integrate the velocity function over the interval [0, π]:
s(π) - s(0) = ∫[0, π] v(t) dt
s(π) - s(0) = ∫[0, π] (-1/2 cos(2t) + 1/3 t^3 + 3.5) dt
s(π) - s(0) = [-1/4 sin(2t) + 1/12 t^4 + 3.5t] [0, π]
s(π) - s(0) = [-1/4 sin(2π) + 1/12 π^4 + 3.5π] - [0]
Since sin(2π) = 0, the distance traveled by the particle between t=0 and t=π is:
s(π) = 1/12 π^4 + 3.5π
Therefore, the distance traveled by the particle between t=0 and t=π is 1/12 π^4 + 3.5π meters.
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Google Gemini AI
May 14
The acceleration of the particle is given by:
$$a = \sin 2t \ t^2 \ m/s^2$$
The velocity of the particle at time t can be found by integrating the acceleration:
$$v = \int a \ dt = \int \sin 2t \ t^2 \ dt$$
Which gives
$$v = \frac{-1}{2} \cos 2t \ t^3 + C$$
At time t = 0, the velocity is given as 3 m/s, so we can solve for C:
$$3 = \frac{-1}{2} \cos 2(0) \ (0)^3 + C$$
$$C = 3$$
Therefore, the velocity of the particle at time t is:
$$v = \frac{-1}{2} \cos 2t \ t^3 + 3$$
The distance traveled by the particle between t=0 and t=π can be found by integrating the velocity:
$$d = \int_{0}^{\pi} v \ dt = \int_{0}^{\pi} \left(\frac{-1}{2} \cos 2t \ t^3 + 3\right) \ dt$$
Which gives
$$d = \left[-\frac{1}{8} \sin 2t \ t^4 + 3t\right]_{0}^{\pi}$$
$$d = \left[-\frac{1}{8} \sin 2(\pi) \ (\pi)^4 + 3(\pi)\right] - \left[-\frac{1}{8} \sin 2(0) \ (0)^4 + 3(0)\right]$$
$$d = \frac{3\pi}{2}$$
Therefore, the distance traveled by the particle between t=0 and t=π is 3π/2 meters.
$$a = \sin 2t \ t^2 \ m/s^2$$
The velocity of the particle at time t can be found by integrating the acceleration:
$$v = \int a \ dt = \int \sin 2t \ t^2 \ dt$$
Which gives
$$v = \frac{-1}{2} \cos 2t \ t^3 + C$$
At time t = 0, the velocity is given as 3 m/s, so we can solve for C:
$$3 = \frac{-1}{2} \cos 2(0) \ (0)^3 + C$$
$$C = 3$$
Therefore, the velocity of the particle at time t is:
$$v = \frac{-1}{2} \cos 2t \ t^3 + 3$$
The distance traveled by the particle between t=0 and t=π can be found by integrating the velocity:
$$d = \int_{0}^{\pi} v \ dt = \int_{0}^{\pi} \left(\frac{-1}{2} \cos 2t \ t^3 + 3\right) \ dt$$
Which gives
$$d = \left[-\frac{1}{8} \sin 2t \ t^4 + 3t\right]_{0}^{\pi}$$
$$d = \left[-\frac{1}{8} \sin 2(\pi) \ (\pi)^4 + 3(\pi)\right] - \left[-\frac{1}{8} \sin 2(0) \ (0)^4 + 3(0)\right]$$
$$d = \frac{3\pi}{2}$$
Therefore, the distance traveled by the particle between t=0 and t=π is 3π/2 meters.
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