Math  /  Calculus

QuestionProblem 12.46 The acceleration of a sled can be prescribed to have one of the following forms: a=β1t,a=β2ta=\beta_{1} \sqrt{t}, a=\beta_{2} t, and a=β3t2a=\beta_{3} t^{2}, where tt is time expressed in seconds, β1=1 m/s5/2,β2=1 m/s3\beta_{1}=1 \mathrm{~m} / \mathrm{s}^{5 / 2}, \beta_{2}=1 \mathrm{~m} / \mathrm{s}^{3}, and β3=1 m/s4\beta_{3}=1 \mathrm{~m} / \mathrm{s}^{4}. The sled starts from rest at t=0t=0. Determine which of the three cases allows the sled to cover the largest distance in 1 s . In addition, determine the distance covered for the case in question.
Figure P12.45 and P12.46 Answm

Studdy Solution
The case with the largest distance is a=β1ta = \beta_1 \sqrt{t}, and the distance covered in 1 second is: x(1)=415m x(1) = \frac{4}{15} \, \text{m}

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