A car accelerates uniformly from rest to a speed of 40 mi/h in 12.0 secs find A) the distance the car traveled during this time and B) the constant accelaration of the car
a) Displacement of an object as a function of time: delta x =1/2(v0+v)T
B) Velocity as a function of discplacment = v^2= v0^2 + 2a delta x
The Attempt at a Solution
A) delta x = 1/2(0 + 20 m/s) 12
answer for part a = delta x = distance =120 m is this right?
B)(20 m/s)^2= 0^2 + 2a(120m)
400 m^2/s^2= 240ma
cross cancellation = answer for part b constant accelaration= 1.666666667= approx 2 m/s^2
Is this right?