Question on Newton's Second Law

Question

A motorist has a reaction time of 0.600 s. While driving at 20.0 m/s, he sees a child run suddenly onto the road, 40.0 m in front of his car. If the mass of the car is 1000 kg and the average horizontal force supplied during braking is 8000 N, determine whether he will be able to stop in time to avoid hitting the child?

Reaction Braking

Answer 1

Initially, since his reaction time is 0.600 s, he travels a distance of (0.600)(20.0) = 12 meters before any slowing begins.

To determine the deceleration, we need to use F = ma, or, solving for "a", we have a = F/m = -8000/1000 = -8m/s²

Using the kinematic equation V_f² = V_i² + 2ax where V_f = 0, V_i = 20, and a = -8, we get:

0² = 20² + 2(-8)x

0 = 400 - 16x

16x = 400

x = 400/16 = 25 m

The total distance the car travels then is 25 + 12 = 37 so, yes, the car stops before hitting the child who is 40 m ahead.

Answer 2

Solution:

First you need to determine the acceleration of the moving car. From Newton second law:

∑F = ma, where the force F = 8000 N = 8000 kg-m/s² is the only acting force with no friction present, m is the mass = 1000 kg, and a is the acceleration in m/s²

rearranging, a = F /m = 8000 kg-m/s² / 1000 kg = 8 m/s²

Using the formula from kinematics V_f² - V_i² = 2 a X, where X can be considered the braking distance needed to stop the car after the brakes are applied, Here The final velocity V_f = 0

The formula becomes 0 - V_i² = 2 a X, solving for X and noting that the acceleration needed is actually a negative deceleration value in the equation = - V_i² / 2 (- a) , and X = - ( 20 m/s)² / 2 ( - 8 m/s²) ,

and X = - 400 m²/s² / 2( - 8 m/s²) = 25 m

The total distance required to stop is the distance traveled during the reaction time =

V₀ t = 20 m/s ( 0.6 s) = 12 m plus the distance traveled during the braking time X,

The total distance needed to stop the car is X_tot = V₀t + (V₀)²/ 2 a = 12 m + 25 m = 37 m < 40 m

Therefore the car will be able to stop before hitting the child