An object uniformly accelerates at a rate of 1.20m/s^2 east. While accelerating at this rate, the object is displaced 470.4m east in a time of 28 s. What velocity did this object reach in this time ?
I used Vf=Vi+at
which will be Vf=(1.20)(28)
but i feel this isn't the way to solve it
please help
Thanks!
Vf^2=Vi^2 +2ad
so the question is, what is Vi?
consider vf=vi+at
Now you have two equations, two unknowns, Vf, and Vi.
I would solve for Vi in the second equation, then put that in the first equation and solve for Vf
You are on the right track! To calculate the final velocity (Vf) of an object, you can use the equation Vf = Vi + at, where Vi represents the initial velocity, a represents the acceleration, and t represents the time.
In this case, the initial velocity is not given, so we can assume it to be zero (Vi = 0). Therefore, the equation simplifies to Vf = at.
Given that the acceleration is 1.20 m/s^2 and the time is 28 s, you can substitute these values into the equation to find the final velocity:
Vf = (1.20 m/s^2) * (28 s)
Vf = 33.6 m/s
So, the object reaches a final velocity of 33.6 m/s east in this time.
Remember that it is crucial to pay attention to the direction while working with velocities. In this case, the direction is specified as east, indicating a positive value.