Question

1a. A 747 airplane starts from rest and reaches a takeoff velocity of 78.0 m/s due west in 5.50 s. Assuming a constant acceleration, how far has it traveled before takeoff? Please explain.

b. Use the component method of vector addition to find the resultant of the following two vectors (magnitude and angle): (show the calculations) A=21.0 km, 26 degree south of west B=38.0km, 30 degrees east of south

Answer 1

1a. using first equation of motion first we will find the constant acceleration by which the airplane starts from rest,

V = U + a*t

where V and U are the final and initial velocities respectively and t is the time taken between start and the takeoff of airplane.

78 = 0 + a*5.5

solving for a,

a = 14.18 m/s²

Now using third equation of motion,

V² - U² = 2*a*s

where s is the distance traveled and a is the constant acceleration.

78² - 0 = 2*14.18*s

solving for s,

s = 214.52 m

Thus, before takeoff the airplane is traveled 214.52 m.

b. firstly we will divide vector A and B into the horizontal and vertical components.

A = 21 km , 26° South of West

A forms the angle with horizontal is (180+26 = 206°),

Horizontal component of A,

Ax = A * Cos ß

Ax = 21 * Cos (206)

Ax = -18.87 km

Vertical component of A,

Ay = A * Sin ß

Ay = 21 * Sin 206

Ay = -9.205 km

B = 38 km, 30° east of south

angle formed between Horizontal and B,

∅ = 270 + 30 = 300°

Horizontal component of B,

Bx = B * Cos ∅

Bx = 38 * Cos 300

Bx = 19 km

Vertical component of B,

By = B * Sin ∅

By = 38 * Sin 300

By = -32.908 km

Now the components of resultant vector of A + B is,

Horizontal component:

ABx = Ax + Bx

ABx = -18.87 + 19

ABx = 0.13 km

Vertical Component:

ABy = Ay + By

ABy = -9.205 - 32.908

ABy = -42.113 km

Magnitude of resultant of vector A+B,

AB = √(ABx² + ABy²)

AB = √(0.13)² + (-42.113)²

AB = 42.114 Km

direction of resultant vector,

theta = arctan (ABy/ABx)

theta = arctan (-42.113/0.14)

theta = -89.82°

Thus the direction of the resultant is (360 - 89.82 = 270.18°), which is 0.18° east of south.

Magnitude = 42.114 km

Direction = 0.18° east of south.