An object is at rest at $(3, 0, 2)$ $(3 ,0 ,2 )$ and constantly accelerates at a rate of $\frac{7}{5} \frac{m}{s^{2}}$ $7/5 m/s^2$ as it moves to point B. If point B is at $(5, 7, 7)$ $(5 ,7 ,7 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.

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Answer 1 · 2017-08-06T06:22:50

$t \approx 3.55 s$ $t ~~ 3.55 s$

${\to s}_{i} = 3 ˆ i + 2 ˆ k$ $vec s_i = 3hati + 2hatk$

${\to s}_{f} = 5 ˆ i + 7 ˆ j + 7 ˆ k$ $vec s_f = 5hati + 7hatj + 7hatk$

$\to s = {\to s}_{f} - {\to s}_{i}$ $vec s = vec s_f - vec s_i$

$\to s = 2 ˆ i + 7 ˆ j + 5 ˆ k$ $vec s = 2hati + 7hatj+ 5hatk$

$∣ ∣ \to s ∣ ∣ = \sqrt{2^{2} + 7^{2} + 5^{2}}$ $|vec s| = sqrt(2^2+7^2+5^2)$

$∣ ∣ \to s ∣ ∣ = \sqrt{78}$ $|vec s| = sqrt(78)$

$∣ ∣ \to s ∣ ∣ = 8.83 m$ $|vec s| = 8.83 m$

$∣ ∣ \to a ∣ ∣ = \frac{7}{5} m \cdot s^{2}$ $|vec a| = 7/5 m*s^2$

$∣ ∣ \to u ∣ ∣ = 0$ $|vec u| = 0$

$∣ ∣ \to s ∣ ∣ = ∣ ∣ \to u ∣ ∣ t + \frac{1}{2} (∣ ∣ \to a ∣ ∣ t^{2})$ $|vec s| = |vec u|t + 1/2(|vec a|t^2)$

$\Rightarrow 8.83 = 0 \cdot t + \frac{1}{2} \cdot \frac{7}{5} \cdot t^{2}$ $rArr 8.83 = 0*t + 1/2*7/5*t^2$

$\Rightarrow 8.83 = \frac{7}{10} t^{2}$ $rArr 8.83 = 7/10t^2$

$\Rightarrow \frac{88.3}{7} = t^{2}$ $rArr 88.3/7 = t^2$

$\Rightarrow t \approx 3.55 s$ $rArr t~~3.55 s$

An object is at rest at (3,0,2)(3 ,0 ,2 ) and constantly accelerates at a rate of 75ms27/5 m/s^2 as it moves to point B. If point B is at (5,7,7)(5 ,7 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.