Question

One day you are driving your friends around town and you drive quickly around a corner without slowing down. You are going at a constant speed of 20.0 m/s ( about 40.0 mph) to the north when you initiate the turn to the right. 2.00 seconds later, you nish the turn and are now traveling to the east, always at constant speed of 20.0 m/s. What was the magnitude of the acceleration experienced by your unfortunate passengers, in terms of g (9.81 m/s2 ), during the turn?

Answer 1

Magnitude of acceleration experienced by the passengers in the car is 1.6g

Step-by-step 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

⇒
`(\pi R)/(2* 2) =20`

R = 80/π

Now,

acceleration ,a = v² / r

a =
`(20^2)/(80/ \pi)`

`a = (400)/(80/ \pi)`

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

Answer 2

1.60 g

Step-by-step explanation:

From the attached file below:

we can deduce that:

`v = v_x =v_y = 20 \ m/s \\t = 2s`

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:

`r = (4x)/(2 \pi)\\\\r = (4*40)/(2 \pi)`

r = 25.5 m

The centripetal acceleration can be expressed as:

`a = (v^2)/(r)`

so;

`a = ((20 \ m/s^2))/(25.5 \ m)`

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:

`a' = (a)/(g) g`

`a' = (15.7 \m/s ^2)/(9.81 \ m/s^2)\ g`

`a' = 1.60 \ g`

∴ The magnitude of the acceleration experienced by your unfortunate passengers during the turn = 1.60 g