Question

A particle starts moving with acceleration 2m/s distance travelled by it in 5th half second is: A1.25 m b 2.25 m c 6.25 m d 30.25 mAnswer is b.but how?

Answer 1

Distance travelled in 5th half second = Distance travelled in 2.5 seconds - Distance travelled in 2 seconds
= 0.5a (t₁² - t₂²)
= 0.5 * 2 * (2.5² - 2²)
= 2.25m

Answer 2

Answer:

Yes the answer is option b. 2.25 m which is solved using instantaneous distant formula.

Explanation:

The particles moves with a given acceleration, so the displacement of the particle in 5th half second will be given by the formula of instantaneous displacement.

Given is the acceleration a = 2 $m/s^{2}$

Time t = 5th half sec =$\frac{5}{2} sec$ = 2.5 sec as $t_{1}$

and $t_{2}$= 5th half sec – one half sec = 4th half sec = $\frac{4}{2}$ = 2 sec

From the instantaneous displacement formula, we have –  

$\bold{\begin{aligned} S_{n}^{t h} &=u+\frac{1}{2} a\left(t_{1}^{2}-t_{2}^{2}\right)} \\ \bold{S_{\frac{5}{2}}^{t h} &=u+\frac{1}{2} a\left(t_{1}^{2}-t_{2}^{2}\right) \end{aligned}}$

$S_{\frac{5}{2}}^{t h}=0+\frac{1}{2} a\left(t_{1}^{2}-t_{2}^{2}\right)$

$\begin{aligned} S_{\frac{5}{2}}^{t h}&=\frac{1}{2} a\left(t_{1}^{2}-t_{2}^{2}\right) \\ S_{\frac{5}{2}}^{t h} &=\frac{1}{2} 2\left(2.5^{2}-2^{2}\right) \end{aligned}$

$\begin{aligned} S_{\frac{5}{2}}^{t h}=&\left(2.5^{2}-2^{2}\right) \\ S_{\frac{5}{2}}^{t h}&=6.25-4 \\ S_{\frac{5}{2}}^{t h}&=2.25 m \end{aligned}$

The displacement in $\bold{5_{th}}$ half second be 2.25 m.