Asked by Katie
a car starts from rest and accelerates for 7.8s with an acceleration of 2.3 m/s squared. how far does it travel
Answers
Answered by
bobpursley
Vf^2=Vi^2+2ad solve for d
where vf= at
Memorize that equation.
where vf= at
Memorize that equation.
Answered by
Ray
using the kinematic equation:
displacement=initial velocity*time+ .5(acceleration*time^2)
delta x= 0*7.8+.5(2.3*7.8^2)
delta x= .5(2.3*7.8^2)
delta x= .5(2.3*60.84)
delta x= 69.966 m
with sig figs your answer would be 70 m
hope this helps!
displacement=initial velocity*time+ .5(acceleration*time^2)
delta x= 0*7.8+.5(2.3*7.8^2)
delta x= .5(2.3*7.8^2)
delta x= .5(2.3*60.84)
delta x= 69.966 m
with sig figs your answer would be 70 m
hope this helps!
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