I found the position and it came out to be 1.959 but I get stuck on part B and C

Question

I found the position and it came out to be 1.959 but I get stuck on part B and C. Can someone help me please?:) A particle moves along the x axis according to the equation x = 1.99 + 3.09t ? 1.00t2, where x is in meters and t is in seconds. (a) Find the position of the particle at t = 3.10 s. =1.959 (b) Find its velocity at t = 3.10 s. ? The instantaneous velocity at a particular time can be obtained by taking the time derivative of the position versus time function and then putting in the specific value of t. m/s (c) Find its acceleration at t = 3.10 s. ?

Explanation / Answer

a)x = 1.99 + 3.09t - 1.00t^2 position at x=3.10 x=1.99+9.6-9.61=2m b)dx/dt=3.09-2.00t=v =3.09-6.2 =-3.11m/s that is 3.11m/s in the opposite direction c)dv/dt=-2.00m/s^2=acceleration

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