## MCAT Question A Day - 2/19/14 - Answer!

**A 2.0-kg body is acted on by a 10-N force. If the body is initially at rest, what will be its velocity after 5 seconds?**

**A.**5.0 m/s

**B.**10 m/s

**C.**20 m/s

**D.**25 m/s

*The correct answer is (D). From Newton’s second law, F 5 ma, so the acceleration is:*

*a = F/m = 10 N / 2.0 kg = 10 kg m s*

^{-2 }/ 2.0 kg = 5.0 m/s^{2}

*The velocity is v = v*

_{0}+ at where v_{0}= 0 so that:

*v = at = (5.0 m/s*

^{2})(5s) = 25 m/s*Want more questions? - Click here to check out our MCAT Question A Day archive*