BEARINGS Question 12 (SSCE 1994 May/June)
(a) A village P is 10km from a lorry station, Q, on
a bearing 065º. Another village R, is 8km from
Q on a bearing 155º. Calculate
(i) the distance of R from P to the nearest
kilometer
(ii) the bearing of R from P to the nearest degree
(b) M is a village on PR such that QM is
perpendicular to PR. Find the distance of M
from P to the nearest kilometer.

a. (i) the distance of R from P is 13 Km to the nearest  kilometer

(ii) the bearing of R from P = 360 - (38 + 25) = 297° to the nearest degree

b. the distance of m from P is 11 Km to the nearest kilometre

Explanation:

a) A triangle PQR is formed. Q = 90°,  p = 8 km; r = 10 km; distance of R from P is q is to be found.

(i) Using the cosine rule: q² = p² + r² - 2prCosQ

q² = 8² + 10² - 2 * 8 * 10 * Cos90

q² = 64 + 100 + 0

q² = 164

q = 13 Km  to the nearest kilometre

Therefore, the distance of R from P is 13 Km to the nearest  kilometer

(ii) the bearing of R from P

The angle at P is found using the formula Cos P = (q² + r² - p²)/2qr

Cos P = 13² + 10² - 8²/2 * 13 *10

Cos P = 0.7884

P = Cos⁻¹ 0.7884

P = 38°

Therefore, the bearing of R from P = 360 - (38 + 25) = 297° to the nearest degree

Note : 25° is alternate (Northwest) to 65°at P

b) A right-angled triangle  QMP is formed

using the trigonometrical ratios; cos Θ = adjacent/hypotenuse

where the hypotenuse side = 10 km, adjacent side = distance of M from P, x

cos P = x/10

x = cos 38 * 10

x = 11 Km to the nearest kilometre

Therefore, the distance of m from P is 11 Km to the nearest kilometre

it’s previous models   because when i picked that son of a gun’s answer it was wrong look for previous models because they might make it a different letter  

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied