Question

The position of a particle as it moves along the x axis is given by x = 15e^-2t m, where t is in s. What is the acceleration of the particle at t = 1.0 s?

Answer 1

Explanation:

यज्ञरहव, जब छठदयडडतझचठतखमबदठडथझभतड

Answer 2

SOLUTION

GIVEN

The position of a particle as it moves along the x axis is given by

$\displaystyle \sf{ x = 15 {e}^{ - 2t} } \: \: m$

where t is in s.

TO DETERMINE

The acceleration of the particle at t = 1.0 s

EVALUATION

Here the position of a particle as it moves along the x axis is given by

$\displaystyle \sf{ x = 15 {e}^{ - 2t} }$

Differentiating both sides with respect to t we get

$\displaystyle \sf{ Velocity = v= \frac{dx}{dt} = - 30 {e}^{ - 2t} }$

Again Differentiating both sides with respect to t we get

$\displaystyle \sf{ Acceleration = a = \frac{ {d}^{2}x }{d {t}^{2} } = 60 {e}^{ - 2t} }$

Now we have

$\displaystyle \sf{ \frac{ {d}^{2}x }{d {t}^{2} } \bigg|_{t = 1.0} = 60 {e}^{ - 2 \times 1} }$

$\displaystyle \sf{ \implies \frac{ {d}^{2}x }{d {t}^{2} } \bigg|_{t = 1.0} = 60 {e}^{ - 2} }$

$\displaystyle \sf{ \implies \frac{ {d}^{2}x }{d {t}^{2} } \bigg|_{t = 1.0} = \frac{60}{ {e}^{2} } }$

$\displaystyle \sf{ \implies \frac{ {d}^{2}x }{d {t}^{2} } \bigg|_{t = 1.0} = \frac{60}{ {(2.72)}^{2} } }$

$\displaystyle \sf{ \implies \frac{ {d}^{2}x }{d {t}^{2} } \bigg|_{t = 1.0} = 8.12 }$

FINAL ANSWER

The acceleration of the particle at t = 1.0 s

= 8.12 m/s²

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