On the second day of Christmas, Calvin gave to me

You need to remember the 3 kinematics equations.

Here, initial velocity $(u)$ = $10m/s$

acceleration $(a)$ = $-2m/s^{2}$

final velocity $(v)$ = $0$

We need to find the displacement $s$ .

So, apply the third kinematics equation:

$v^{2} = u^{2} + 2as$

$0 = 10^{2} - 2 \times 2 \times s$

$100 = 4s$

$s = 25$ meters

That's the answer!

Initial velocity $= 10 ms^{-1}$

Final velocity $= 0 ms^{-1}$

Acceleration $= -2 ms^{-2}$

$$v^2 - u^2 = 2as$$ $$0^2 - 10^2 = 2(-2)s$$ $$-100 = -4s$$ $$s = 25$$

We have initial velocity $(u)=10 m/s$ and acceleration $(a)=(-2) m/s^{2}$ . Since it is said that she is decelerating, she will travel the distance(say 's' m) till her velocity becomes zero, i.e, final velocity $(v)=0$ .

We now use here the motion equation $v^{2}=u^{2}+2as$ . Putting the given values in the equation, we get---

$0^{2}=10^{2}+2\times (-2) \times s \implies 0=100-4s \implies 4s=100 \implies s=\boxed{25}$

Thus we finally get the distance covered $= s = \boxed{25}$

During the first second, the speed of the turtle is greater than 8 m/s. This means it does more than 8 meters. During the 2nd second, its speed is greater than 6 m/s, so it will go on for more than 6 meters. 8+6=14 meters, and this exclude three of the possible answers :P

It takes 5 seconds for Turt to completely stop. If Turt did not decelerate he would have travelled a distance of 5s x 10m/s = 50m , but he did decelerate uniformly, thus he travelled a distance of 50m / 2 = 25m.

$v^{2}$ = $u^{2}$ - $2as$

$0$ = $10^{2}$ - $2 \times 2 \times s$

$100$ = $4s$

$s$ = $\boxed{25 meters}$

v =0 u = 10m/s a= -2 m/ sec. square

by 3rd eq of motion , v (sq) - u (sq) = 2as we get by substituting the values , s = 25m s = distance traveled by the Turt is 25m

Well, using Torricelli We have : V² = Vo²-2.a.∆s = > 0 = 100-4.∆s => ∆s = 25m.

Vt = Vo + at ---> Vt = 0 + 2t ---> 10 = 2t ---> t = 5. We know if S = Vo . t + 1/2 . a . t^2 ---> S = 0 + 1/2 . a . t^2 = 1/2 . 2 .5 ^2 = 25. Answer : 25