can you find out the 3 decimal places in final answer of A and B? thank you 14:08A brainly.comBRAINLYSearch...Answer:a) The additional time required for the truck tostop is 8.5 secondsb) The additional distance traveled by thetruck is 230.05 ftExplanation:Since the acceleration is constant, the averagespeed is:(final speed - initial speed) / 2 = 0.75 v0Since travelling at this speed for 8.5 secondscauses the vehicle to travel 690 ft, we can solvefor vo:0.75v0 * 8.5 = 690vO = 108.24 ft/sThe speed after 8.5 seconds is: 108.24 /2 =54.12 ft/sWe can now use the following equation to solvefor acceleration:v2 – u? = 2 * a * s54.122 – 108.24² = 2 * a * 690a = -6.367 m/s^2Additional time taken to decelerate: 54.12/6.367= 8.5 secondsTotal distance traveled:v² – u² = 2 * a * s0 - 108 24^2 = 2-6 36 S 14:08A brainly.comBRAINLYSearch...speed is:(final speed - initial speed) / 2 = 0.75 v0Since travelling at this speed for 8.5 secondscauses the vehicle to travel 690 ft, we can solvefor vo:0.75v0 * 8.5 = 690vO = 108.24 ft/sThe speed after 8.5 seconds is: 108.24 / 2 =54.12 ft/sWe can now use the following equation to solvefor acceleration:v² – u? = 2 * a * s54.122 – 108.24² = 2 * a * 690a = -6.367 m/s^2Additional time taken to decelerate: 54.12/6.367= 8.5 secondsTotal distance traveled:v2 – u? = 2 * a * sO - 108.24^2 = 2 * (-6.367) * ssolving for s we get total distance traveled =920.05 ftAdditional Distance Traveled: 920.05 - 690 =230.05 ftocabanga44 and 1 more users found this answer helpful

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can you find out the 3 decimal places in final answer of A and B?

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter6: Applications Of Newton's Laws

Section: Chapter Questions

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Scalars and Vectors

Scalars and Vectors are the classification of physical quantities that are used to denote various terms in physics.

Question

can you find out the 3 decimal places in final answer of A and B? thank you

14:08
A brainly.com
BRAINLY
Search...
Answer:
a) The additional time required for the truck to
stop is 8.5 seconds
b) The additional distance traveled by the
truck is 230.05 ft
Explanation:
Since the acceleration is constant, the average
speed is:
(final speed - initial speed) / 2 = 0.75 v0
Since travelling at this speed for 8.5 seconds
causes the vehicle to travel 690 ft, we can solve
for vo:
0.75v0 * 8.5 = 690
vO = 108.24 ft/s
The speed after 8.5 seconds is: 108.24 /2 =
54.12 ft/s
We can now use the following equation to solve
for acceleration:
v2 – u? = 2 * a * s
54.122 – 108.24² = 2 * a * 690
a = -6.367 m/s^2
Additional time taken to decelerate: 54.12/6.367
= 8.5 seconds
Total distance traveled:
v² – u² = 2 * a * s
0 - 108 24^2 = 2-6 36 S

14:08
A brainly.com
BRAINLY
Search...
speed is:
(final speed - initial speed) / 2 = 0.75 v0
Since travelling at this speed for 8.5 seconds
causes the vehicle to travel 690 ft, we can solve
for vo:
0.75v0 * 8.5 = 690
vO = 108.24 ft/s
The speed after 8.5 seconds is: 108.24 / 2 =
54.12 ft/s
We can now use the following equation to solve
for acceleration:
v² – u? = 2 * a * s
54.122 – 108.24² = 2 * a * 690
a = -6.367 m/s^2
Additional time taken to decelerate: 54.12/6.367
= 8.5 seconds
Total distance traveled:
v2 – u? = 2 * a * s
O - 108.24^2 = 2 * (-6.367) * s
solving for s we get total distance traveled =
920.05 ft
Additional Distance Traveled: 920.05 - 690 =
230.05 ft
ocabanga44 and 1 more users found this answer helpful

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