Question

A body moving along a straight line with a velocity 20m/s undergoes an acceleration of -4 m/s2 .After 2s its speed will be

Answer 1

Answer:

12 m/s

Explanation:

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

• Initial velocity (u) = 20 m/s
• Acceleration (a) = – 4 m/s²
• Time taken (t) = 2 s

We have been asked to calculate its speed after 2 seconds.

Here, we can easily find its speed after 2 seconds by using the first equation of motion. By using the 1st equation of motion :

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

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

Substitute the values in the formula.

$\longrightarrow \sf{\quad { v = 20 + (-4)2}}$

$\longrightarrow \sf{\quad { v = 20 + (-8)}}$

$\longrightarrow \sf{\quad { v = 20 - 8}}$

$\longrightarrow \quad\underline{\boxed { \pmb{\frak{ v = 12 \; m\: s}}^{ \pmb{\frak{ -1}}} }}$

Therefore, speed of it after 2 seconds will be 12 m/s.

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Learn More :

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