Question

A particle accelerates from rest at a constant rate for some time and attains a velocity of 8m/s. Afterwards it deaccelerates with the constant rate and comes to rest .If the total time taken is 4sec, what is the distance travelled?​

Answer 1

Given:-

• Initial velocity ,u = 0m/s

• Final velocity ,v = 8m/s

• Time taken ,t = 4

To Find:-

• Distance travelled ,s

Solution:-

Firstly we calculate the acceleration of the particles.

Using 1st Equation of Motion

• v = u +at

where,

v is the final velocity

a is the acceleration

u is the initial velocity

t is the time taken

Substitute the value we get

→ 8 = 0 + a ×4

→ 8 = a×4

→ a = 8/4

→ a = 2m/s²

Acceleration is 2m/s²

Now, Calculating the Distance travelled.

Using 3rd equation of motion

• v² = u² +2as

Substitute the value we get

→ 8² = 0² + 2×2 × s

→ 64 = 0 + 4×s

→ 64 = 4×s

→ s = 64/4

→ s = 16 m

Therefore, the distance travelled by the particles is 16 metres.

Answer 2

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A particle accelerates from rest at a constant rate for some time and attains a velocity of 8m/s. Afterwards it deaccelerates with the constant rate and comes to rest .If the total time taken is 4sec, what is the distance travelled?

${ { \underbrace{ \mathbb{ \red{GiVeN\ }}}}}$

• $Initial \:velocity(u) =0m/s$
• $Final\:velocity (v) =8m/s$
• $Time\: taken (t) =4s$

${ { \underbrace{ \mathbb{ \red{To\:PrOvE\ }}}}}$

$Distance \:travelled \:of \:a \:particle$

$\star\underbrace{\mathtt\red{⫷❥ᴀ} \mathtt\green{n }\mathtt\blue{ s} \mathtt\purple{W}\mathtt\orange{e} \mathtt\pink{R⫸}}\star\:$

• ${\boxed {\boxed {v=u+at}}}$
• $v²=u²+2as$
• $s=ut+\frac{1}{2} at²$

$v=Final\:velocity$

$u=Initial\:velocity$

$a=acceleration$

$t=Time\:taken$

$now\:substitute\:the \:values$

$8=0+a(4)$

$8=a\times 4$

$a= \frac{8}{4}$

${\boxed {\boxed {a=2m/s²}}}$

________________________________

• $v=u+at$
• ${\boxed {\boxed {v²=u²+2as}}}$
• $s=ut+\frac{1}{2} at²$

$now\:substitute\:the\:values$

$8²=0²+2(2)(s)$

$64=0+4\times s$

$64=4\times s$

$s=\frac{64}{4}$

${\boxed {\boxed {s=m}}}$

$\therefore the\:distance \:travelled \:by\:the \:particles \:is\:{\boxed {\boxed {16m}}}$

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