Question

You are driving along a highway at a steady 60 mph (88 ft/s) when you see an accident ahead and slam on the brakes. What constant deceleration is required to stop your car in 242 ft? To find out, carry out the following steps:

Solve the initial value problem differential equation: d^2s/dt^2
a) Solve the initial value problem differential equation
b) Determine the required deceleration
c) Analyze the impact of different initial conditions
d) Assess the braking distance under varying speeds

Answer 1

The required constant deceleration to stop a car in 242 feet when driving at 60 mph is 31.68 ft/s². By applying kinematic equations, we can assess the braking distance under varying speeds and understand the impact of different initial conditions.

Step-by-step explanation:

To answer the question of what constant deceleration is required to stop a car in 242 feet when driving at a steady 60 mph (which is 88 ft/s), we can use the kinematic equation:

v2 = u2 + 2as, where:

• v is the final velocity (0 ft/s because the car stops),
• u is the initial velocity (88 ft/s),
• a is the acceleration (deceleration in this case), and
• s is the displacement (242 ft).

Solving for a, we get:

0 = (88)2 + 2a(242)

a = -(88)2 / (2 * 242)

a = -31.68 ft/s2 (negative sign indicates deceleration)

The required deceleration is 31.68 ft/s2. Now let's assess the braking distance under varying speeds. For example, driving at exactly 50 mph (which is 73.3 ft/s), and using the same kinematic equation with u = 73.3 ft/s and a = same as calculated before, we can find the new stopping distance.

s = -(u2) / (2a)

s = -(73.3)2 / (2 * (-31.68))

s = 85.093 ft

This demonstrates how initial conditions such as speed influence the stopping distance. The impact of different initial conditions can be critical in assessing safety scenarios.