Now let\'s assume that we know x(3s) = 2m, v(3s) = 1 m/s, and a (3s) = -0.5 m/s^

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Now let's assume that we know x(3s) = 2m, v(3s) = 1 m/s, and a (3s) = -0.5 m/s^2, where a(t) is the acceleration (i.e., a(t) = d/dt v(t)). Explain why knowledge of the position at time t_0 doesn't help as understand the velocity at time t t_0. Explain why knowledge of the velocity at time t_0 doesn't help us understand tin- acceleration at time t t_0. Explain why knowledge of the acceleration at time t_0 does actually help us understand how the velocity might change as t varies around t_0. Using the assumption above, justify the approximation: a(t) 0.5 m/s^2. Now, given your approximation for the acceleration, you know that the velocity must satisfy d/dt v(t) = a(t). That is, the velocity is the antiderivative of the acceleration. Justify the approximation: v(t) v_0 + (t - 3s) middot 0.5 m/s^2 for some constant v_0 (in meters/second). Use your knowledge of the velocity (remember that v(3s) = 1 m/s) to explain why the constant v_0 must in fact be equal to 1 m/s. Thus, show that: v(t) 1 m/s + (t - 3s) middot 0.5 m/s^2. Now, given your approximation for the velocity, you know that the position must satisfy d/dt x (t) = v (t). That is, the position (as a function) must be the antiderivative of the velocity. Explain why: x(t) x_0 + (t - 3s) middot 1 m/s + 1/2 (t - 3s)^2 middot 0.5 m/s^2 for some unknown constant x_0 (in meters). Use your knowledge of the position (remember that x(3s) = 2m) to explain why the constant x_0 must in fact be equal to 2m. Construct what is called the 'second-order taylor-series' approximation x(t) 2m + (t - 3s) middot 1 m/s + 1/2 (t - 3s)^2 middot 0.5 m/s^2.

Explanation / Answer

1. Velocity is the derivative function of position vector.

Just knowing the postiion at a particular time, will not help us to find velocity until we have the velocity as a function of position. if position r(t) then velocity v(t) = r'(t) and then we can put t=t0.

Therefore just knowing the value of position at any particular time will not help us understand velocity.

2. Similarly as above, acceleration is derivative function of velocity.

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