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19 days ago

13

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?

2 answers:

Ostrovityanka [2.2K]19 days ago

6 0

Result:

1.60 g

Elucidation:

Based on the attached document:

we can infer that:

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

The distance covered in 2 seconds will be:

x = vt

x = 20 m/s × 2 s

x = 40 m

The segment corresponds to a quarter of a circle with radius r,

therefore, if 2 πr = 4 x

Then the radius (r) can be calculated as:

$r = \frac{4x}{2 \pi}\\\\r = \frac{4*40}{2 \pi}$

r = 25.5 m

Centripetal acceleration can be expressed as:

$a = \frac{v^2}{r}$

thus;

$a = \frac{(20 \ m/s^2)}{25.5 \ m}$

a = 15.7 m/s²

The acceleration magnitude suffered by your passengers in relation to the acceleration due to gravity can be calculated using:

$a' = \frac{a}{g} g$

$a' = \frac{15.7 \m/s ^2}{9.81 \ m/s^2}\ g$

$a' = 1.60 \ g$

∴ The acceleration magnitude experienced by your passengers while turning = 1.60 g

kicyunya [2.2K]19 days ago

4 0

Result:

The acceleration magnitude felt by the car's passengers is 1.6g

Explanation:

Given the following:

Gravitational acceleration is defined as g = 9.81 m/s2

The car's speed is V = 20 m/s

The time spent navigating the curve is T = 2 sec

Radius of the turn = R

Distance traveled, d = θ r = (π/2)R

Also,

d/t = v

⇒ $\frac{\pi R}{2\times 2} =20$

R = 80/π

Now,

Acceleration can be calculated as a = v² / r

a = $\frac{20^2}{80/ \pi}$

$a = \frac{400}{80/ \pi}$

a = 400 / 25.46

a = 15.708m/s²

Expressed in terms of g:

a = 15.708 / 9.8g

a = 1.60g

As a result, the acceleration magnitude felt by the car's passengers is 1.6g

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