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Question

https://s2.lite.msu.edu/enc/65/d3fdafc75c99f7b44529b75d4c6b3ef80a264b4c859e5c7358b1caf4a3aa44428578f5d91c400efd30b68056afd459553ec53d399a54c03df3b25fba673d05addcf8bedee7c0b5e0b94d246a722ecdf42821c6a024b5b482?symb=%2fenc%2f130%2fdaca2fe9cc8574acbb2945aa521778ebe31de24687debce570abc37ecc98864e8082b4ba66ae5ec9251af31610138705ac00d8428951626792199b4305645c217c28bda2402ff2f7bcb98baed3e17e1ba7b3b448101b7b357ac8109856ee6b4225d47b0adf817cfb5653e5fcd3460cfa254da315210b16e96c672b0a20491e28d51c098e491e1a4a28b91432d4f4935cf22a5b1fdfed2121ef44c4e4515eb8f381bca996aa2f3a1b

The graph shows the speed of a car traveling in a straight line as a function of time.

The value of Vc is 4.20 m/s and the value of Vd is 5.30 m/s. Calculate the distance traveled by the car from a time of 1.20 to 5.20 seconds.

Explanation / Answer

vc =4.2

vd =5.3

t1= 1.2 s

t2 =5.2 s

a= 5.3-4.2/5.2-1.2 =0.367

v^2-u^2 = 2as

s= v^2-u^2 /2a

5.3^2 -4.2^2 / 2*0.367

=14.2370572207

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