# Distance formula using initial velocity and acceleration

We know $$S=ut+1/2at^2$$. Now, let's say my initial velocity is $$u=10\text { m/s}$$ and acceleration $$a=5\text{ m/s}^2$$.

In first second distance traveled $$S^{t=1}=10\text { m}$$ as my speed is $$10\text{ m/s}$$ and time is $$1 \text{ s}$$.

In next second my speed will be $$15m/s$$ as $$a=5m/s^2$$and distance travelled will be $$S^{_t=2}=S^{_t=1} + 15=25m$$ as distance traveled in 2nd second is $$15m$$.

Now when I apply the same case in equation $$S=ut+1/2at^2$$ where $$u=10m/s$$, $$a=5m/s^2$$ and $$t=2s$$ I get $$S=30m$$

Where Iam I getting wrong?

• Suppose you're travelling at a constant speed of 10 m/s, then after the first second you travel a distance of 10 m. Right. But if you're accelerating then during that first second your speed will increase above 10 m/s and therefore you will travel more than 10 m. Mar 9, 2016 at 16:50
• Hello and Welcome to Stack Exchange. Note that we are not a homework help-site and check-my-work questions are generally off-topic. That said, the problem is that your first idea (first second/second second) doesn't take into account that the speed is NOT 10 m/2 for the whole second, but gradually increases. Mar 9, 2016 at 16:52