Question

Suppose that another car is moving with a speed of 18.5 m/s when the brakes are applied so as to slow the car down at a rate of -2.85 m/s.

a. What will be the speed of this car 3.55 seconds after the brakes are applied? (Round to 1 decimal) m/s
b. How far will this car move during this 3.55 second period? (Round to 1 decimal) meters
c. How long will it take for this car to stop? (Round to 1 decimal) seconds
d. How far will this car move from the time that the brakes are applied until the car comes to a stop? (Round to 1 decimal) meters

Answer 1

The student's question involves physics concepts of deceleration, speed, and distance calculations for a car when brakes are applied. It includes determining the time and distance it takes for a car to come to a stop after deceleration.

Step-by-step explanation:

The question relates to the deceleration of a car and involves finding the final speed after a certain time, the distance covered in that time, and the total time and distance to stop the car.

a. To find the speed of the car after 3.55 seconds when brakes are applied with an acceleration of -2.85 m/s2, you use the equation: v = u + at. Where:
v is the final velocity,
u is the initial velocity (18.5 m/s),
a is the acceleration (-2.85 m/s2),
t is the time (3.55 s).

b. To determine the distance the car moves in 3.55 seconds, use the equation: s = ut + (1/2)at2.

c. The time to stop the car is found by setting the final velocity to 0 m/s in the equation v = u + at and solving for t.

d. The total distance to stop the car is found using the equation s = (v2 - u2)/(2a), with v equal to 0 m/s.