CALC? The position of a squirrel running in a park is given by (a) What are vx(t) and vy(t), the x- and y-components of the velocity of the squirrel, as functions of time? (b) At t = 5.00 s, how far is the squirrel from its initial position? (c) At t = 5.00 s, what are the magnitude and direction of the squirrel’s velocity?

Solution 4E Step 1 : In this question, we need to find The x and y components of velocity of the squirrel ,that is v (t) axd v (t) y Distance travelled by the squirrel from its initial position at time t=5s The magnitude and direction of velocity of squirrel at time t= 5 s Step 2 : Let us consider the condition given r = [(0.280m/s)t + (0.0360 m/s )t ] i + (0.0190 m/s ) t j 3 dr We know that v = dt So we shall differentiate the above condition with respect to time dx 2 dt = (0.280 m/s) + 2(0.0360 m/s )t dx = 0.280 m/s + 0.072 m/s t 2 dt Similarly we have dy = 3(0.0190 m/s )t 2 dt Hence we have X-component as dx = 0.280 m/s + 0.072 m/s t 2 dt dy 3 2 Y-component as dt = 3(0.0190 m/s )t Step 3 : To find distance travelled at time t=5s r = [(0.280m/s)t + (0.0360 m/s )t ] i + (0.0190 m/s ) t j 2 r(x) = [(0.280m/s) × 5 + (0.0360 m/s )5 ] i r(x) = [(0.280m/s) × 5 + (0.0360 m/s )5 ] i r(x) = 1.4 m/s + 0.9 m/s = 2.3 m Similarly for y component 3 3 r(y) = (0.0190 m/s ) t j r(y) = (0.0190 m/s ) × 5 j 3 r(y) = (0.0190 m/s ) × 125 j 3 r(y) = 2.375 m/s We can get initial distance as 2 2 | | r(x) + r(y) Substituting the values 2 2 | | 2.3 m + 2.375 m | | 10.930 m = 3.3 m Hence the distance travelled by squirrel from initial position is 3.3 m