Question

Consider a particle initially moving with a velocity of 5 $m{s}^{-1}$ starts decelerating at a constant rate of 2 $m{s}^{-2}$.
a. Determine the time at which the particle becomes stationary.
b. Find the distance travelled in the second second.
c. Find the distance travelled

Answer 1

a. Here u = 5$m{s}^{-1}$, a = -2 $m{s}^{-2}$, v= 0, t = ?

Using v = u + at, where

0 = 5 - 2t $\Rightarrow$ t = 2.5s

b. Here u = 5$m{s}^{-1}$, a = -2 $m{s}^{-2}$, n = 2.

Using $x_n=u+\frac{a}{2}(2n-1)=5-\frac{2}{2}\left[2\right(2)-1)]=2m$

c. Here, if we use the above formula, we will get $x_n$ = 0, but in reality, it is not zero. This formula is not applicable for the third second because velocity becomes zero in the third second, i.e., at t = 2.5 s. The particle has a turning point at t = 2.5 s. We have to indirectly calculate the distance travelled in this particular second. That is, we have to determine the distance travelled, between 2 $\le$ t $\le$ 2.5 and 2.5$\le$t$\le$3, and then add the two.

Displacement of the particle at t = 2.5 s is

$x_{2.5}=\frac{u^2}{2a}=\frac{(5{)}^2}{2\left(2\right)}$ = 6.25m

Due to symmetry, the displacement of the particle at t = 2 s and t = 3 s are same, i.e.,

$x_3=x_2=5\left(2\right)+\left(1/2\right)(-2)(2{)}^2$ = 6m

Thus, the distance travelled in the third second is

$x=(x_{2.5}-x_2)+(x_{2.5}-x_3)$

= ( 6.25 - 6 ) + ( 6.25 - 6 ) = 0.5m