Question

particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i = 6.00î m/s. (a) Find the vector position of the particle at any time t (where t is measured in seconds). ( t î + t2 ĵ) m (b) Find the velocity of the particle at any time t. ( î + t ĵ) m/s (c) Continued in next problem with new numbers. Find the coordinates of the particle at t = 5.00 s. x = m y = m (d) Find the speed of the particle at t = 5.00 s. m/s

Answer 1

Answer:

(a)$6t\hat{i} + t^2\hat{j}$

(b)$6\hat{i} + 2t\hat{j}$

(c)x = 30m, y = 25m

(d)11.66m/s

Explanation:

While position has a formula of $at^2/2 + vt$, velocity has a formula of a*t.

(a) the vector position of the particle at any time t is

$s(t) = v_it\hat{i} + a_jt^2/2\hat{j} = 6t\hat{i} + t^2\hat{j}$

(b) the vector velocity of the particle at any time t is

$v(t) = v_i\hat{i} + a_jt\hat{j} = 6\hat{i} + 2t\hat{j}$

(c) at t = 5 s

$s(5) = 6*5\hat{i} + 5^2\hat{j} = 30\hat{i} + 25\hat{j}$

So x = 30 m and y = 25m

(d) $v(5) = 6\hat{i} + 2*5\hat{j} = 6\hat{i} + 10\hat{j}$

with $v_i = 6 m/s, v_j = 10m/s$So the speed quantity is

$\sqrt{6^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136} = 11.66 m/s$