Answer:
1.60 g
Explanation:
From the attached file below:
we can deduce that:
[tex]v = v_x =v_y = 20 \ m/s \\t = 2s[/tex]
The distance traveled by 2 s will be:
x = vt
x = 20 m/s × 2 s
x = 40 m
The length is quarter of the circle with radius r,
so; if 2 πr = 4 x
Then radius (r) will be:
[tex]r = \frac{4x}{2 \pi}\\\\r = \frac{4*40}{2 \pi}[/tex]
r = 25.5 m
The centripetal acceleration can be expressed as:
[tex]a = \frac{v^2}{r}[/tex]
so;
[tex]a = \frac{(20 \ m/s^2)}{25.5 \ m}[/tex]
a = 15.7 m/s²
the magnitude of the acceleration experienced by your unfortunate passengers in terms of acceleration due to gravity is then determined by the equation:
[tex]a' = \frac{a}{g} g[/tex]
[tex]a' = \frac{15.7 \m/s ^2}{9.81 \ m/s^2}\ g[/tex]
[tex]a' = 1.60 \ g[/tex]
∴ The magnitude of the acceleration experienced by your unfortunate passengers during the turn = 1.60 g
Answer:
Magnitude of acceleration experienced by the passengers in the car is 1.6g
Explanation:
Given that,
Gravitational acceleration = g = 9.81 m/s2
Speed of the car = V = 20 m/s
Time taken to drive around the curve = T = 2 sec
Radius of the curve = R
distance cover, d = θ r = (π/2)R
Also ,
d/t = v
⇒[tex]\frac{\pi R}{2\times 2} =20[/tex]
R = 80/π
Now,
acceleration ,a = v² / r
a = [tex]\frac{20^2}{80/ \pi}[/tex]
[tex]a = \frac{400}{80/ \pi}[/tex]
a = 400 / 25.46
a = 15.708m/s²
in term of g
a = 15.708 / 9.8g
a = 1.60g
Therefore, Magnitude of acceleration experienced by the passengers in the car is 1.6g
