Question

To practice Problem-Solving Strategy 2.1 Motion with constant acceleration You are driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. Your reaction time before stepping on the brakes is 0.50 s , and the maximum deceleration of your car is 10 m/s2 . How much distance is between you and the deer when you come to a stop
a. How much distance is between you and the deer when you come to a stop?
b. What is the maximum speed you could have and still not hit the deer?

Answer 1

Answer:

a) $\Delta s=5\ m$ is the distance between deer and the vehicle

b) $u'=22.36\ m.s^{-1}$ is the maximum speed the driver can be at and still not hit the deer.

Explanation:

Given:

• initial speed of driving, $u=20\ m.s^{-1}$

• distance of deer from the vehicle, $x=35\ m$

• reaction time taken to step onto the brakes, $t'=0.5\ s$

• maximum deceleration of the car, $a_m=-10\ m.s^{-2}$

a)

Now the distance travelled after application of the brakes till the vehicle stops:

$v^2=u^2+2a_m.s$

(assuming that the brakes are applied with maximum acceleration)

where:

$s=$ displacement of the vehicle after braking till it stops

$v=$ final velocity of the vehicle = 0 (stops)

putting the values:

$0^2=20^2-2\times 10\times s$

$s=20\ m$

Now before the application of the brakes 0.5 second is taken to react and the vehicle travels during this time as well.

So, distance covered before applying the brakes:

$s'=u.t'$

$s'=20\times 0.5$

$s'=10\ m$

The distance between the deer and the vehicle:

$\Delta s=x-(s+s')$

$\Delta s=35-(20+10)$

$\Delta s=5\ m$

b)

The maximum speed the driver can have with the vehicle and still not hit the deer is given as:

$v^2=u'^2+2. a_m.(x-s')$

because s' is the distance covered before braking during the reaction time.

$0^2=u'^2-2\times 10\times (35-10)$

$u'=22.36\ m.s^{-1}$ is the maximum speed the driver can be at and still not hit the deer.