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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 .

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