Challenge Problem (Extra Credit) A car with bad shocks has a mass of 1500 kg. Before you go for a drive with three of your friends you notice that the car sinks a distance of 6.0 cm when all four of you get in the car. You estimate that the four of you together have a mass of [11] kg. As you are driving down the highway at 65 mph you notice that the car is starting to bounce up and down with large amplitude. You realize that there is a periodic series of small bumps and dips in the road that is driving the bouncing. What is the distance (in meters, to two significant figures) between adjacent bumps on the road, assuming that damping by the shocks is negligible

167.354 m

Explanation:

We are given;

The mass of the car with bad shock;

m = 1500 kg

The distance at which the car sinks; x =

6 cm = 6 × 10^(−2) m

The total mass of 4 people; m_t = 11 kg

The total speed in the highway; V = 65

mph = 29.058 m/s

The spring's constant can be calculated from the formula;

F = Kx

F is also equal to mg.

Thus;

m_t × g = Kx

K = (m_t × g)/x

K = (11 × 9.81)/(6 × 10^(−2))

K = 1798.5 N/m

Mass of car and four people;m_(c+t) = 1500 + 11 = 1511 kg

Thus, the period cam be calculated from the formula;

T = 2π√((m_c+t)/k)

T = 2π√(1511/1798.5)

T = 5.759 s

the distance between adjacent bumps is calculated from;

Velocity = distance/time

Distance = velocity x time

Distance = 29.058 × 5.759

Distance = 167.354 m

does anyone know or what?

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.