# Chapter 3 - Vectors and Coordinates Systems - Exercises and Problems: 23

(a) At t = 0: The particle's distance from the origin is zero. At t = 2 s: The particle's distance from the origin is 25.6 m At t = 5 s: The particle's distance from the origin is 160 m (b) $v(t) = [(10.0\hat{i}+8.0\hat{j})~t] ~m/s$ (c) At t = 0: The particle's speed is 0 At t = 2 s: The particle's speed is 25.6 m/s At t = 5 s: The particle's speed is 64.0 m/s

#### Work Step by Step

$\vec{r} = (5.0\hat{i}+4.0\hat{j})~t^2 ~m$ (a) At t= 0: $\vec{r} = (5.0\hat{i}+4.0\hat{j})~(0)^2 ~m$ $\vec{r} = 0$ The particle's distance from the origin is zero. At t = 2 s: $\vec{r} = (5.0\hat{i}+4.0\hat{j})~(2~s)^2 ~m$ $\vec{r} = (20.0\hat{i}+16.0\hat{j})~m$ We can find the distance from the origin. $r = \sqrt{(20.0~m)^2+(16.0~m)^2}$ $r = 25.6~m$ The particle's distance from the origin is 25.6 m At t = 5 s: $\vec{r} = (5.0\hat{i}+4.0\hat{j})~(5~s)^2 ~m$ $\vec{r} = (125\hat{i}+100\hat{j})~m$ We can find the distance from the origin. $r = \sqrt{(125~m)^2+(100~m)^2}$ $r = 160~m$ The particle's distance from the origin is 160 m (b) $\vec{r} = (5.0\hat{i}+4.0\hat{j})~t^2 ~m$ $v(t) = \frac{dr}{dt}$ $v(t) = [(5.0\hat{i}+4.0\hat{j})~2t] ~m/s$ $v(t) = [(10.0\hat{i}+8.0\hat{j})~t] ~m/s$ (c) At t = 0: $v = (10.0\hat{i}+8.0\hat{j})~(0) ~m/s$ $v = 0$ At t = 2 s: $v = (10.0\hat{i}+8.0\hat{j})~(2~s)~m/s$ $v = (20.0\hat{i}+16.0\hat{j})~m/s$ We can find the speed. $v = \sqrt{(20.0~m/s)^2+(16.0~m/s)^2}$ $v = 25.6~m/s$ The particle's speed is 25.6 m/s At t = 5 s: $v = (10.0\hat{i}+8.0\hat{j})~(5~s)~m/s$ $v = (50.0\hat{i}+40.0\hat{j})~m/s$ We can find the speed. $v = \sqrt{(50.0~m/s)^2+(40.0~m/s)^2}$ $v = 64.0~m/s$ The particle's speed is 64.0 m/s

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