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

the entire statement is true and correct.  

there's no diagram, nothing is illustrated, and no question is asked.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft