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An object moving on the x axis with a constant acceleration increases its x coordinate
by 108 m in a time of 3.68 s and has a velocity
of 20 m/s at the end of this time.
Determine the acceleration of the object
during this motion. Answer in units of m/s^2

V. Harshit - , 12 Years ago
Grade 11
anser 1 Answers
Arun Kumar
Hello Student,

V_f-V_i=at=a*3.68
V_f^2-V^2=2as=2*108*a
=>V_f+V_i=at=2*108*a/a*3.68=216/3.68
soV_f=216/3.68+a*3.68=20
=>a=(20-(216/3.68))/3.68=-10.5
So the particle was actually accelerating.

Thanks & Regards
Arun Kumar
Askiitians Faculty
IIT Delhi
Last Activity: 11 Years ago
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