Question

U=4,a= 3m/ secs2, t= ? , S= 8

Answer 1

Answer:

1.33 seconds

Explanation:

As per the provided information in the given question, we have :

• Initial velocity, u = 4 m/s
• Acceleration, a = 3 m/s²
• Distance travelled, s = 8 m

We are asked to calculate the time taken. Let's assume the time taken as t.

According to the 3rd equation of motion,

$\longrightarrow \quad \boxed{\pmb{\sf {v^2 - u^2 = 2as}} }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

➝ v² – (4)² = 2 × 3 × 8

➝ v² – 16 = 48

➝ v² = 48 + 16

➝ v² = 64

➝ v = √64

➝ v = 8 m/s

Now, by using the first equation of motion :

$\longrightarrow \quad \boxed{\pmb{\sf {v = u + at}} }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

➝ 8 = 4 + 3t

➝ 8 – 4 = 3t

➝ 4 = 3t

➝ 4 ÷ 3 = t

➝ 1.33 seconds = t

∴ Time taken is 1.33 seconds.

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More Information :

Equations of motion :

$\boxed{ \begin{array}{cc} \pmb{\sf{ \quad \: v = u + at \quad}} \\ \\ \pmb{\sf{ \quad \: s= ut + \cfrac{1}{2}a{t}^{2} \quad \: } } \\ \\ \pmb{\sf{ \quad \: {v}^{2} - {u}^{2} = 2as \quad \:}}\end{array}}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time
• s denotes distance