Calculating Vehicle Deceleration: Time & Distance

[ajayespinoza]

Homework Statement

A 4200lb vehicle is cruising down the highway at 65mph when the driver steps on the brake applying a braking force of 2,850lbs
Determine the time to stop, assuming uniform deceleration
Distance traveled during the deceleration
Velocity of the vehicle after 3.2 sec's of braking

Homework Equations

V = (p2-p1)/time

The Attempt at a Solution

I currently do not have any attempts at a solution other than
4200-2850 then divide that by 65?

 

[tiny-tim]

A 4200lb vehicle is cruising down the highway at 65mph when the driver steps on the brake applying a braking force of 2,850lbs
Determine the time to stop, assuming uniform deceleration
Distance traveled during the deceleration
Velocity of the vehicle after 3.2 sec's of braking

Hi ajayespinoza!

Hint: Start by using good ol' Newton's second law to find the deceleration.

 

[baba89]

I would approach this problem by first converting all units to the standard unit system, such as meters and seconds. Then, I would use the equations of motion, specifically the one that relates velocity, time, and acceleration (V = V0 + at), to solve for the unknown variables.

To determine the time to stop, we need to find the acceleration of the vehicle. This can be done by dividing the braking force by the mass of the vehicle (2850lbs/4200lbs = 0.68 m/s^2). Then, using the equation V = V0 + at, we can solve for the time to stop (assuming the initial velocity, V0, is 65mph or 29.06 m/s).

t = (V - V0)/a
t = (0 - 29.06 m/s)/-0.68 m/s^2
t = 42.74 seconds

Therefore, it would take approximately 42.74 seconds for the vehicle to come to a complete stop.

To determine the distance traveled during the deceleration, we can use the equation x = x0 + V0t + 1/2at^2, where x0 is the initial position and x is the final position. Since the vehicle is initially traveling at a constant velocity, x0 can be set to 0. Plugging in the known values, we get:

x = 0 + (29.06 m/s)(42.74 s) + 1/2(-0.68 m/s^2)(42.74 s)^2
x = 620.9 meters

Therefore, the vehicle would travel approximately 620.9 meters during the deceleration.

To determine the velocity of the vehicle after 3.2 seconds of braking, we can again use the equation V = V0 + at, but this time with t = 3.2 seconds.

V = (29.06 m/s) + (-0.68 m/s^2)(3.2 s)
V = 26.5 m/s

Therefore, the velocity of the vehicle after 3.2 seconds of braking would be approximately 26.5 m/s.