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Given :
Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south.
To Find :
The magnitude and direction of
a) A + B .
b) A - B.
Solution :
Let , direction in north is given by +j and east is given by +i .
So ,
and 
Now , A + B is given by :


Direction of A+B is 45° north of west .
Also , for A-B :


Direction of A-B is 45° south of west .
( When two vector of same magnitude which are perpendicular to each other are added or subtracted the resultant is always 45° from each of them)
Hence , this is the required solution .
Answer:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.
Answer:
100N
Explanation:
Newton's third law of motion
For every action, there is an equal and opposite reaction.
Therefore 100N of force is exerted by the crate on student as a reaction to his action