Sue is driving at 32 m/s and enters a single-lane tunnel. She observes a slow moving van 160 m ahead traveling at 5.10 m/s. Sue applies her brakes but can only accelerate at -1.30 m/s2 because the road is wet. How far does she end up in the tunnel?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

The initial speed `V_0 ` is `32 m / s ` while the acceleration `a ` is `- 1.3 m / s^2 , ` which means the speed will change by the rule

`V_S = V_S ( t ) = V_0 + a t = 32 - 1.3 t`

until the speed becomes zero. The distance from the the starting point will be `D_S = 32t - 1/2*1.3t^2 .`

Also, the van is moving away with the speed `V_v = 5.1 m / s , ` so the speed of convergence is

`V_c = V_S ( t ) - V_v = 26.9 - 1.3 t .`

The initial distance between the vehicles is `D_0 = 160 m ,` therefore the distance at time `t ` will be

`D = D ( t ) = 160 - 26.9 t + 1/2 * 1.3 t^2 .`

(We get this by integration.) Collision occurs when `D ( t ) = 0 ` for the smallest `t gt 0 ` or, in other words, at

`t_1 = 1/1.3 (26.9 - sqrt(26.9^2-160*2.6)) approx 7.2 ( s ) . `

We should also check that `V_S ( t_1 ) gt 0 ` (it is true).

The distance into the tunnel will be `D_S ( t_1 ) approx 196.7 ( m ) .`

Last Updated by eNotes Editorial on