Arace car is driving on the bonneville salt flats in utah. it starts out at at rest, 40.0 m east (+x) and 32.0 m north (+y) of a camping facility, which we will consider to be the origin. the race car proceeds to accelerate at 8.50 m/s 2 at 63.0º north of east. write the unit-vector equations that describe the car’s (a) position r ? ? , (b) velocity v ? ? , and (c) acceleration a ? ? . (d) how long does it take for the car to reach a speed of 200 km/hr? (e) where is the car when it reaches this speed?

a) r = (40.0 m + 1.93 m/s² · t²) i + (32.0 m + 3.79  m/s² · t²) j

b) v = (3.86 m/s² i +7.57 m/s² j) · t

c) a = 3.86 m/s² i + 7.57 m/s² j

d) It takes 6.54 s for the car to reach a speed of 200 km/h

e) The position of the car when it reaches this speed is r = 123 m i + 194 m j

Explanation:

Please, see the attached figure for a better understanding of the problem.

c)The components of the vector acceleration are ax and ay (see the figure). Then:

a = (ax, ay)

Using trigonometry :

(cos α = adjacent/ hypotenuse and sin α = opposite/hypotenuse)

a = (8.50 m/s² · cos 63°, 8.50 m/s² · sin 63°)

a = 3.86 m/s² i + 7.57 m/s² j

b) The vector velocity will be the acceleration multiplied by the time (v = a · t because the car starts from rest, so v0 = 0)

v = (3.86 m/s² i +7.57 m/s² j) · t

a) The position is calculated as:

position = initial position + v0 · t + 1/2 ·a· t²

The x-component of the vector position "r" will be:

x = 40.0 m + 1/2 · 3.86 m/s² · t²      (v0 = 0)

x = 40.0 m + 1.93 m/s² · t²

The y-component will be:

y = 32.0 m + 1/2 · 7.57 m/s² · t²

y = 32.0 m + 3.79  m/s² · t²

Then, the position vector will be:

r = (40.0 m + 1.93 m/s² · t²) i + (32.0 m + 3.79  m/s² · t²) j

d) The magnitude of the velocity is 200 km/h (55.6 m/s). Then:

Solving for t:

14.9 m²/s⁴ · t² + 57.3 m²/s⁴ · t² = (55.6 m/s)²

72.2 m²/s⁴ · t² = 3.09 × 10³ m²/s²

t² = 3.09 × 10³ m²/s² / 72.2 m²/s⁴

t = 6.54 s

e)Now, we just have to replace t = 6.54 in the equation of the position of the car:

r = (40.0 m + 1.93 m/s² · t²) i + (32.0 m + 3.79  m/s² · t²) j

r(t=6.54 s) = (40.0 m + 1.93 m/s² · (6.54 s)²) i + (32.0 m + 3.79  m/s² · (6.54 s)²) j

r = 123 m i + 194 m j

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered