Question

Please answer the numerical with full explanation below.............

Answer 1

GiveN :

• Initial Velocity (u) = 6 m/s
• Final Velocity (v) = 16 m/s
• Time (t) = 10 sec

To FinD :

• Acceleration
• Distance Traveled

SolutioN :

Use formula for acceleration :

$\large \dashrightarrow {\boxed{\sf{a \: = \: \dfrac{v \: - \: u}{t}}}} \\ \\ \dashrightarrow \tt{a \: = \: \dfrac{16 \: - \: 6}{10}} \\ \\ \dashrightarrow \tt{a \: = \: \dfrac{10}{10}} \\ \\ \dashrightarrow \tt{a \: = \: 1} \\ \\ \underline{\boxed{\bf{Acceleration \: is \: 1 \: ms^{-2}}}}$

$\rule{200}{2}$

Use 3rd equation of motion

$\large \dashrightarrow {\boxed{\sf{v^2 \: - \: u^2 \: = \: 2as}}} \\ \\ \dashrightarrow \tt{16^2 \: - \: 6^2 \: = \: 2 \: \times \: 1 \: \times \: s} \\ \\ \dashrightarrow \tt{256 \: - \: 36 \: = \: 2s} \\ \\ \dashrightarrow \tt{220 \: = \: 2s} \\ \\ \dashrightarrow \tt{s \: = \: \dfrac{220}{2}} \\ \\ \dashrightarrow \tt{s \: = \: 110} \\ \\ \underline{\boxed{\textsf{\textbf{Distance \: Travelled \: is \: 110 \: m}}}}$

Answer 2

Solution :

⏭ Given:

✏ Initial velocity of car = 6mps

✏ Final velocity of car = 16mps

✏ Time interval = 10s

⏭ To Find:

• Acceleration of car
• Distance covered by car in that time interval

⏭ Concept:

✏ Since, this is uniform acceleratory motion...First, we have to calculate acceleration of body after that we can find out distance covered by body.

⏭ Formula:

✏ Relation between change in velocity, time interval and acceleration is given by

$\bigstar \: \underline{ \boxed{ \bold{ \sf{ \pink{ \large{a = \dfrac{v - u}{t}}}}}}}$

✏ Formula of distance covered by body in terms of initial velocity, acceleration and time interval is given by

$\bigstar \underline{ \boxed{ \bold{ \sf{ \purple{ \large{d = ut + \dfrac{1}{2} a {t}^{2} }}}}}}$

⏭ Calculation:

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• Acceleration

$\twoheadrightarrow \sf \: a = \dfrac{16 - 6}{10} \\ \\ \twoheadrightarrow \sf \: a = \dfrac{10}{10} \\ \\ \twoheadrightarrow \: \boxed{ \sf{ \red{a = 1 \: m {s}^{ - 2}}}} \: \gray{ \surd}$

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• Distance

$\dashrightarrow \sf \: d = (6 \times 10) + { \huge{(}}\dfrac{1}{2} \times 1 \times {10}^{2}{ \huge{)}} \\ \\ \dashrightarrow \sf \: d = 60 + 50 \\ \\ \dashrightarrow \: \boxed{ \sf{ \orange{d = 110 \: m}}} \: \gray{ \surd}$

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