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An object, which is at the origin at time t=0t=0, has initial velocity v0=(14.0i^7.0j^)m/s\overrightarrow{\mathbf{v}}_0=(-14.0 \hat{\mathbf{i}}-7.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} and constant acceleration a=(6.0i^+3.0j^)m/s2\overrightarrow{\mathbf{a}}=(6.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^2. Determine the position r\overrightarrow{\mathbf{r}} where the object comes to rest (momentarily).

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Answered 2 years ago
Answered 2 years ago
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The knowns are:

  • v0=14.0  m/s i^7.0  m/s j^\vec{\bm{v}}_0 = -14.0\;\mathrm{m/s}\ \bm{\hat{i}} - 7.0\;\mathrm{m/s}\ \bm{\hat{j}}
  • a=(6.0 i^+3.0 j^) m/s²\vec{\bm{a}} = (6.0\ \bm{\hat{i}} + 3.0\ \hat{\bm{j}})\ \mathrm{m/s²}

Our goal is to calculate the position vector at the moment when the particle comes to rest.

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