Question

A particle initially at rest moves with an acceleration 5 metre per second square for 5 seconds find the distance travelled in 4 seconds 5 seconds and fifth second
answer it ​

Answer 1

Answer:

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Answer:

The distance travelled in 4 seconds is 40 m.

The distance travelled in 5 seconds is 62.5m.

The distance travelled in 5th second is 22.5 m.

Explanation:

Given:

Initial velocity (u) = 0

Acceleration (a) = 5 m/s²

$\underline{\pink{\tt{The \: distance \: travelled \: in \: 4 s}}}$

The distance travelled by the particle in t = 4 s

Let the distance travelled in 4 seconds by the particle be '$S_{1}$'

So,

By using the equation of motion,

$S_{1}$ = ut + $\frac{1}{2} a{t}^{2}$

$= 0 \times 4 + \frac{1}{2} \times 5 \times {4}^{2}$

= 40 m

Therefore, the distance travelled in 4 seconds by the particle is 40 m.

$\underline{\pink{\tt{The \: distance \: travelled \: in \: 5 s}}}$

The distance travelled by the particle in t = 4 s

Let the distance travelled in 4 seconds by the particle be '$S_{2}$'

So,

By using the equation of motion,

$S_{2}$ = ut + $\frac{1}{2} a{t}^{2}$

$= 0 \times 5 + \frac{1}{2} \times 5 \times {5}^{2}$

= 62.5 m

Therefore, the distance travelled in 5 seconds by the particle is 62.5 m.

$\underline{\pink{\tt{The \: distance \: travelled \: in \:5th \: second }}}$

Answer 2

u=0

a=5 ms  −2

 

As it undergoes acceleration for 5 s, the distance travelled in 4 s is due to an accelerated motion. t=4 s

∴s=ut+  2 1  at  2  

=(0×4)+(  2 1×5×4  2  )

=40 m