Math / Calculus

Question(II) A car drives straight down toward the bottom of a valley and up the other side on a road whose bottom has a radius of curvature of 125 m . At the very bottom, the normal force on the driver is twice his weight. At what speed was the car traveling?

Studdy Solution

STEP 1

1. The car is moving in a circular path at the bottom of the valley.
2. The radius of curvature of the path is $125 \, \text{m}$ .
3. The normal force on the driver is twice his weight at the bottom of the valley.
4. The weight of the driver is given by $mg$ , where $m$ is the mass of the driver and $g$ is the acceleration due to gravity ( $9.8 \, \text{m/s}^2$ ).

STEP 2

1. Understand the forces acting on the driver at the bottom of the valley.
2. Set up the equation for the net force in terms of centripetal force.
3. Solve for the speed of the car.

STEP 3

Understand the forces acting on the driver at the bottom of the valley:
- The normal force ( $N$ ) is acting upwards. - The gravitational force ( $mg$ ) is acting downwards. - At the bottom of the valley, the normal force is twice the weight of the driver, i.e., $N = 2mg$ .

STEP 4

Set up the equation for the net force in terms of centripetal force:
- The net force towards the center of the circular path is the centripetal force, which is given by $F_c = \frac{mv^2}{r}$ . - At the bottom of the valley, the net force is $N - mg = \frac{mv^2}{r}$ .
Substitute $N = 2mg$ into the equation:
$2mg - mg = \frac{mv^2}{r}$
Simplify the equation:
$mg = \frac{mv^2}{r}$

STEP 5

Solve for the speed of the car:
- Cancel $m$ from both sides of the equation:
$g = \frac{v^2}{r}$
- Solve for $v^2$ :
$v^2 = gr$
- Substitute the known values ( $g = 9.8 \, \text{m/s}^2$ and $r = 125 \, \text{m}$ ):
$v^2 = 9.8 \times 125$
- Calculate $v$ :
$v = \sqrt{9.8 \times 125}$ $v \approx \sqrt{1225}$ $v \approx 35 \, \text{m/s}$
The speed of the car was approximately:
$\boxed{35 \, \text{m/s}}$

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