# Distance formula using initial velocity and acceleration

we know $S=ut+1/2at^2$. Now lets say my initial velocity $u=10m/s$ and acceleration be $a=5m/s^2$.

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

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. – lemon Mar 9 '16 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. – Martin Mar 9 '16 at 16:52