## College Physics (4th Edition)

The correct answer is: (c) three times the distance covered during the first second Let $a$ be the acceleration. We can find the distance covered in the first second: $\Delta x = \frac{1}{2}at^2 = \frac{1}{2}a(1.0)^2 = \frac{a}{2}$ We can find the velocity after $1.0 ~s$: $v_f = v_0+at = 0+(a)(1.0) = a$ We can find the distance covered between $t=1.0~s$ and $t=2.0~s$: $\Delta x = v_0t+\frac{1}{2}at^2 = a(1.0)+ \frac{1}{2}a(1.0)^2 = a+\frac{a}{2} = \frac{3a}{2}$ We can see that the distance covered between $t=1.0~s$ and $t=2.0~s$ is three times the distance covered in the first second.