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A freight train travels at v = 60(1 - { e }^{ -t }) ft/s, where t is the elapsed time in seconds. Determine the distance traveled in three seconds, and the acceleration at this time.

Step-by-step

v = 60(1 – { e }^{ -t }) \\ \int_{ 0 }^{ s } { ds } = \int { v }\space dt = \int_{ 0 }^{ 3 } { 60(1 – { e }^{ -t })dt } \\ s = 60(t + { e }^{ -t })|_{ 0 }^{ 3 } \\ s = 123 \text{ ft } \\ a = \frac { dv } { dt } = 60({ e }^{ -t }) \\ \text{ At } t = 3s \\ a = 60{ e }^{ -3 } = 2.99 \text{ ft }/{ s }^{ 2 }

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