The deceleration(negative acceleration) of an automobile in a full skid on the dr pavement is a little greater than one half "g" or half the acceleration you would experience in free fall. Suppose you are going down the high way and you suddenly decide you need to stop your car. your maximum acceleration is 5.5m/s2 (assumed constant) and you travelling at 26.7m/s.
A. What is the minimum time it will take to stop.?
B. What will your average velocity be while stopping?
C. How much distance will it take you to stop?
A. 10 seconds
To solve this problem, we can use the kinematic equations of motion. Let's break down each part of the question and find the answers step by step.
A. What is the minimum time it will take to stop?
To find the minimum time it will take to stop, we need to use the equation:
v = u + at
Where:
v = final velocity (0 m/s because the car needs to come to a stop)
u = initial velocity (26.7 m/s)
a = acceleration (-5.5 m/s²)
t = time
Rearranging the equation, we have:
t = (v - u) / a
Plugging in the values:
t = (0 - 26.7) / (-5.5)
t = 4.85 seconds
Therefore, it will take approximately 4.85 seconds to stop the car.
B. What will your average velocity be while stopping?
Since we want to find the average velocity, we can use the equation:
average velocity = (initial velocity + final velocity) / 2
Plugging in the values:
average velocity = (26.7 + 0) / 2
average velocity = 13.35 m/s
Therefore, the average velocity while stopping is 13.35 m/s.
C. How much distance will it take you to stop?
To find the distance traveled, we can use the equation:
d = ut + (1/2)at²
Where:
d = distance traveled
u = initial velocity (26.7 m/s)
t = time taken to stop (4.85 seconds)
a = acceleration (-5.5 m/s²)
Plugging in the values:
d = (26.7 × 4.85) + (0.5 × -5.5 × 4.85²)
d = 129.495 + (-67.161)
d = 62.334 meters
Therefore, it will take approximately 62.334 meters to stop the car.
Remember to always double-check the units and make sure they are consistent throughout the calculations.
If you have any further questions, feel free to ask!