Question

A car is stopped at a traffic signal. A second car rear-ends the first car. From the crush of the two vehicles, it is estimated that the second car was traveling at 15 mph when it impacted the first car. If the second car laid down 166 feet of skid marks before the crash took place, how fast was the second car traveling when its brakes were locked up

Answer 1

The question is incomplete. The complete question is :

A car is stopped at a traffic signal. A second car rear-ends the first car. From the crush of the two vehicles, it is estimated that the second car was traveling at 15 mph when it impacted the first car. If the second car laid down 166 feet of skid marks before the crash took place, how fast was the second car traveling when its brakes were locked up? The approach to the intersection where the accident took place is on a 2.5 percent downgrade. Assume the AASHTO recommended deceleration rate of
`$11.2 \ ft/s^2.$`

Solution :

From the equation, the skid distance is given by

`$S=(u_i^2-u_f^2)/(2g\left((a_d)/(g)+h\right))$`

Here,
`$a_d$` = deceleration rate,
`$11.2 \ ft/s^2.$`

h =
`$-(2.5)/(100)$` = -0.025

`$u_i$` = initial speed of the car

`$u_f$` = speed when impacted

Therefore,

`$166=(u_i^2-(15 * 1.467)^2)/(2* 32.2\left((11.2)/(32.2)+0.025\right))$`

`$\Rightarrow u_i^2 = 3935.36$`

`$\Rightarrow u_i = 62.73 \ ft/s$`

or
`$ u_i = (62.73)/(1.467) $`

`$ u_i = 42.76 \ mph $`