Answer:
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:
[tex]|v| = \sqrt{(3.86 m/s^{2} * t)^{2} + (7.57 m/s^{2} * t)^{2}} = 55.6 m/s[/tex]
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
