Question

An ambulance has successive displacement of 6.3km, 45 degrees, 12km south and 4.7km, 30 degrees, south of west a.]Determine the magnitude and deriction of the resultan displacement b.]What is the total distance traveled by ambulance ?

Answer 1

a) The resultant displacement is AD vector. AD = AB + BC + CD = AC + CD

$|AC|^2 = |AB|^2 + |BC|^2 - 2 |AB| |BC| \cos 45 ^\circ = 6.3^2 +12^2 - 2 \cdot 6.3 \cdot 12 \cdot \cos 45 ^\circ = 76.77$

$|AC| = 8.76$ km.

$|AD|^2 = |AC|^2 + |CD|^2 - 2 |AC| |CD| \cos (90^\circ + 30^\circ) = 8.76^2 +4.7^2 - 2 \cdot 8.76 \cdot 4.7 \cdot \cos (120^\circ) = 140$

$|AD| = 11.83$ km.

b) Total distance travelled is 6.3 km + 12 km + 4.7 km = 23 km.

Answer: a) 11.83 km, b) 23km.