A car traveling at 35.0 km/h speeds up to 45.0 km/h in a time of 5.60 s (interva

Question

A car traveling at 35.0 km/h speeds up to 45.0 km/h in a time of 5.60 s (interval A). The same car later speeds up from 65.0 km/h to 75.0 km/h in a time of 5.60 s (interval B). 1) Calculate the magnitude of the constant acceleration for the first interval. (Express your answer to three significant figures.)2) Calculate the magnitude of the constant acceleration for the second interval. (Express your answer to three significant figures.)3) Determine the distance traveled by the car during the first time interval. (Express your answer to three significant figures.)4-Determine the distance traveled by the car during the second time interval. (Express your answer to three significant figures.)

Explanation / Answer

(1) First of all convert the speed in m/s -

v1 = 35 km/h = (35*1000) / (60*60) = 9.72 m/s

like-wise, v2 = 45.0 km/h = 12.5 m/s

t = 5.6 s

use the expression -

v2 = v1 + a1*t

=> 12.5 = 9.72 + a1*5.6

=> a1 = (12.5 - 9.72) / 5.6 = 0.50 m/s^2

(3) Distance travelled by the car in this interval s1 = (v2^2 - v1^2) / 2a1 = (12.5^2 - 9.72^2) / (2*.50) = 61.8 m

(2) v3 = 65 km/h = (65*1000) / (60*60) = 18.06 m/s

like-wise, v2 = 75.0 km/h = 20.8 m/s

t = 5.6 s

use the expression -

v2 = v1 + a1*t

=> 20.8 = 18.06 + a2*5.6

=> a2 = (20.8 - 18.06) / 5.6 = 0.50 m/s^2

(4) Distance travelled by the car in this interval s1 = (20.8^2 - 18.06^2) / 2a1 = (20.8^2 - 18.06^2) / (2*0.50)

= 106.50 m

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