Question

A truck is moving with velocity 20 metre per second if the retardation of 20 metre per second square is produced in the truck, calculate the distance covered before comming to rest. (Please answer Fast)

Answer 1

Given :

• u = 20m/s
• v = 0m/s t
• t = 4s

To Find :

• Distance = ?

Solution :

Using first equation of motion,

• v = u + at

That gives us retardation :

→ 0 = 20 + 4t

→ -20 = 4t

→ t = 5 m/s²

Using second equation of motion ,

→ s = ut + 1/2at²

→ s = 20(4) + 1/2 × (-5) × 4 × 4

→ s = 80 + (-40)

→ s = 80 - 40

→ s = 40m

Hence, distance covered by the truck before coming in rest is 40m.

Answer 2

Given :-

• u = 20 m/s
• v = 0 m/s
• t = 4 s

To find :-

• Find Distance ?

Solution :-

$\star$$\boxed{\frak{First ~equation~ of ~motion~ —}}$$\star$

• $\large\dashrightarrow$v = u + at

$\star$$\boxed{\frak{Substituting ~the~ values :}}$$\star$

• $\large\mapsto$0 = 20 + 4t
• $\large\mapsto$-20 = 4t
• $\large\mapsto$t = 5 m/s²

$\star$$\boxed{\frak{Second ~equation ~of ~motion~ —}}$$\star$

:$\implies$ s = ${\sf{ut \: + \: \frac{1}{2} \: {at}^{2} }}$

$~~~~~$:$\implies$ s = ${\sf{20(4) + \frac{1}{2} \times (-5) \times 4 \times 4}}$

$~~~~~~~~~~$:$\implies$80 + (-40)

$~~~~~~~~~~~~~~~$:$\implies$80 - 40

$~~~~~~~~~~~~~~~~~~~~$:$\implies$$\large{\underline{\boxed{\bf{\frak{\pink{s~=~40m}}}}}}$★

$\large\dag$ Hence,

• Distance covered by the truck before coming is rest is $\large\underline{\rm{40m}}$ $\large\green\checkmark$