Answered

Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

A car comes to a stop six seconds after the driver applies the brakes. While the brakes are on, the following velocities are recorded:
Time since brakes applied (sec) 0 2 4 6
Velocity (ft/s) 87 44 16 0
Give lower and upper estimates (using all of the available data) for the distance the car traveled after the brakes were applied.

Sagot :

Answer:

Explanation:

deceleration during first 2 s = (87 - 44) / 2 =21.5 m /s²

deceleration during next 2 s = (44 - 16) / 2 =14 m /s²

deceleration during last  2 s = (16 - 0) / 2 = 8  m /s²

If we assume the acceleration to be uniform during 3 readings

distance travelled during first 2 s

v² - u² = 2 a s

87² - 44² = 2 x 21.5 x s

s = 7569 - 1936 / 43

= 176 m

distance travelled during next 2 s

44² - 16² = 2 x 14 x s

s =  1936 - 256  / 28

= 60 m

distance travelled during last 2 s

16² - 0² = 2 x 8 x s

s =   256  / 16

= 16 m

Total distance travelled = 176 + 60 + 16 = 252 m

If we assume the acceleration as non- uniform , we first calculate average acceleration as follows .

average acceleration = (87 - 0)/ 6 = 14.5 m /s²

v² - u² = 2 a s

87² - 0² = 2 x 14.5 x s

s = 7569  / 29

= 261  m

Hence lower estimate of distance travelled = 252 m ; upper estimate of distance travelled = 261 m .

Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.