original acceleration is a
1/2 at^2 = 37
v = at
new acceleration is 2a
since v = 2at, new time is (2.3v)/(2a) = 2.3at/2a = 1.15t
new displacement is
1/2 (2a)*(1.15t)^2 = 2*1.15^2 at^2 = 5.29*37 = 195.73 m
An object starts from rest an undergoes a constant acceleration in the positive x direction. After some time it covers a displacement of 37 meters in the positive x direction achieving some final velocity. If the object started from rest with double the original acceleration (acceleration is constant again) and then got of to a final velocity which is a factor of 2.3 larger than the original final velocity, what is the magnitude of the new displacement in meters? Hint: Use an appropriate constant acceleration equation and use proportions.
1 answer
0 answers