Final answer:
The car takes approximately 3.99 seconds to decelerate from 20 m/s to 5 m/s with a force of 8410 N acting to the east. Within this time, the car travels approximately 49.87 meters.
Step-by-step explanation:
The scenario describes a physics problem involving deceleration, force, and displacement of a car. To find the time of deceleration (a), we use the formula force = mass × acceleration. Given that the force to the east is 8410 N and the mass of the car is 2240 kg, the acceleration (a) can be calculated as:
a = Force / Mass = 8410 N / 2240 kg = 3.754 N/kg (or m/s²)
The deceleration is to the west, so the acceleration to the east is -3.754 m/s². Now, we apply the formula v = u+at to find the time (t), where v is the final velocity (5 m/s), u is the initial velocity (20 m/s), and a is the acceleration (-3.754 m/s²). We get:
5 m/s = 20 m/s + (-3.754 m/s²)t
t = (5 m/s - 20 m/s) / (-3.754 m/s²) = 3.99 s
For part (b), to find the distance (d) traveled during deceleration, we can use the formula s = ut + ½at²:
s = 20 m/s × 3.99 s + ½(-3.754 m/s²)(3.99 s)²
s = 79.8 m + ½(-3.754 m/s²)(15.92 s²)
s = 79.8 m - 29.93 m
s = 49.87 m
Answers:
(a) It takes the car approximately 3.99 seconds to decelerate. (b) The car travels approximately 49.87 meters during the deceleration.