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› Part (c) Find the instantaneous velocity, in meters per second, at t = 3.15 s. Vx(13) = -123.0175 Feedback: is available. Vx(13)=-123 X Attempts Remain Part (d) Find the instantaneous acceleration, in meters per square second, at t = 1.3 s. ax(t₁) = 12.78 X Attempts Remain Part (e) Find the instantaneous acceleration, in meters per square second, at t = 2.35 s. Part (f) Find the instantaneous acceleration, in meters per square second, at t = 3.15 s.

› Part (c) Find the instantaneous velocity, in meters per second, at t = 3.15 s. Vx(13) = -123.0175 Feedback: is available. Vx(13)=-123 X Attempts Remain Part (d) Find the instantaneous acceleration, in meters per square second, at t = 1.3 s. ax(t₁) = 12.78 X Attempts Remain Part (e) Find the instantaneous acceleration, in meters per square second, at t = 2.35 s. Part (f) Find the instantaneous acceleration, in meters per square second, at t = 3.15 s.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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please only answer parts c,d,e,f

In a particular Cartesian coordinate system, the y and z-components of the velocity of a point-like object are zero and the x- component varies as given by the following function: vx(t)=2+51-31². where t is in seconds and vx is in meters per second. Part (a) Find the instantaneous velocity, in meters per second, at t = 1.3 s. ✓ Correct! Vx(1₁) = 3.43 Part (b) Find the instantaneous velocity, in meters per second, at t = 2.35 s. Vx(1₂) = -2.8175 Vx(1₂) = -2.818 ✓ Correct! › Part (c) Find the instantaneous velocity, in meters per second, at t = 3.15 s. Vx(13)=-123.0175 Feedback: is available. vx(13)=-123 X Attempts Remain Part (d) Find the instantaneous acceleration, in meters per square second, at t = 1.3 s. a(t₁) 12.78 X Attempts Remain Part (e) Find the instantaneous acceleration, in meters per square second, at t = 2.35 s. Part (f) Find the instantaneous acceleration, in meters per square second, at t = 3.15 s. Part (g) If the particle was at x = 0 at t = 0, find the position, in meters, of the particle at t = 1.3 s. ✓ Correct! x(t₁) = 4.628 Part (h) If the particle was at x = 0 at t = 0, find the position, in meters, of the particle at t = 2.35 s. ✓ Correct! x(+₂) = 5.528
Transcribed Image Text:In a particular Cartesian coordinate system, the y and z-components of the velocity of a point-like object are zero and the x- component varies as given by the following function: vx(t)=2+51-31². where t is in seconds and vx is in meters per second. Part (a) Find the instantaneous velocity, in meters per second, at t = 1.3 s. ✓ Correct! Vx(1₁) = 3.43 Part (b) Find the instantaneous velocity, in meters per second, at t = 2.35 s. Vx(1₂) = -2.8175 Vx(1₂) = -2.818 ✓ Correct! › Part (c) Find the instantaneous velocity, in meters per second, at t = 3.15 s. Vx(13)=-123.0175 Feedback: is available. vx(13)=-123 X Attempts Remain Part (d) Find the instantaneous acceleration, in meters per square second, at t = 1.3 s. a(t₁) 12.78 X Attempts Remain Part (e) Find the instantaneous acceleration, in meters per square second, at t = 2.35 s. Part (f) Find the instantaneous acceleration, in meters per square second, at t = 3.15 s. Part (g) If the particle was at x = 0 at t = 0, find the position, in meters, of the particle at t = 1.3 s. ✓ Correct! x(t₁) = 4.628 Part (h) If the particle was at x = 0 at t = 0, find the position, in meters, of the particle at t = 2.35 s. ✓ Correct! x(+₂) = 5.528
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