The acceleration of a particle along a straight line is defined by a = (2t – 9) m/s^2 , where t is in seconds. At t = 0, s = 1 m and v = 10 m/s. When t = 9 s, determine (a) the particle’s position, (b) the total distance traveled, and (c) the velocity.

Question

The acceleration of a particle along a straight line is defined by a = (2t - 9) m/{ s }^{ 2 } , where t is in seconds. At t = 0, s = 1 m and v = 10 m/s. When t = 9 s, determine (a) the particle’s position, (b) the total distance traveled, and (c) the velocity.

Accepted Answer

[latex]a = 2t - 9 \ \int _{ 10 }^{ v }{ dv = \int _{ 0 }^{ 1 }{ (2t - 9) dt } } \ v -10 = { t }^{ 2 } -9t \ v = { t }^{ 2 } -9t + 10 \ \int_{ 1 }^{ s } ds = \int_{ 0 }^{ t }({ t }^{ 2 } - 9t + 10)dt \ s - 1 = \frac { 1 } { 3 } { t }^{ 3 } - 4.5{ t }^{ 2 } + 10t \ s = \frac { 1 } { 3 } { t }^{ 3 } - 4.5 { t }^{ 2 } + 10 t + 1[/latex]

Note when [latex]v = { t }^{ 2 } - 9 t + 10 = 0[/latex]:

[latex]t = 1.298 s ext{ and } t = 7.701 s[/latex]

When [latex]t = 1.298 s,\quad s = 7.13 m[/latex]

When [latex]t = 7.701 s,\quad s = -36.63 m[/latex]

When [latex]t = 9 s,\quad s = -30.5 m[/latex]

[latex] ext{ (a) } \quad s = -30.5m \ \begin{aligned} ext{ (b) } \quad &{ s }_{ To t } = (7.13 - 1) + 7.13 + 36.63 + (36.63 - 30.50) \ & { s }_{ To t } = 56.0m \end{aligned} \ ext{ (c) } \quad v = 10\space m/s[/latex]

Question : The acceleration of a particle along a straight line is defi...

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