A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 6.70 m/s. What is the x-coordinate of the object when t = 10.0 s?
Added by Brittney W.
Step 1
Given ax(t) = -(0.032 m/s^3)(15.0s - t), integrate to get Vx(t): ∫ax(t) dt = ∫-(0.032)(15.0 - t) dt Vx(t) = -(0.032)(15t - (t^2)/2) + C1 Vx(t) = -0.48t + 0.016t^2 + C1 Show more…
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