Question

Please help in Calculating Acceleration if the car starts from rest and gains a Velocity of 5 m/s in 2 s. Don't forget to find Distance Travelled

Answer 1

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$\red{\underline{{ \bf Question }}}$

• Calculate the acceleration and distance travelled of a car which starts from rest and gains a velocity of 5 m/s in 2 sec.

$\red{\underline{{ \bf Answer }}}$

$\underline {\underline{{\purple{ \sf Given }}}}$

• Initial Velocity (u) = 0 m/s
• Final Velocity (v) = 5 m/s
• Time Taken (t) = 2 sec

$\underline {\underline{{ \purple{\sf To \: Find }}}}$

• Acceleration
• Distance Travelled

$\underline {\underline{{ \green{\sf Calculating \: Acceleration }}}}$

Formula Used :- $\underline{\underline{\boxed{ \pink{\sf v = u + at }}}}$

Substituting Values

☞ $\blue {\sf 5 = 0 + a \times 2 }$

☞ $\blue {\sf 5 = 0 + 2a }$

☞ $\blue {\sf 5 = 2a }$

☞ $\blue {\sf 2a = 5 }$

☞ $\blue {\sf a = \dfrac{5}{2} }$

☞ $\blue {\sf a = 2.5}$

Therefore, acceleration 2.5 m/s²

$\underline {\underline{{ \green{\sf Calculating \: Distance \: Travelled }}}}$

Formula Used :- $\underline{\underline{\boxed{ \pink{\sf{v}^{2} - {u}^{2} = 2as }}}}$

Substituting Values

☞ $\blue {\sf (5)^{2} - (0)^{2} = 2 \times 2.5 \times s }$

☞ $\blue {\sf 25 - 0 = 5s }$

☞ $\blue {\sf 25 = 5s }$

☞ $\blue {\sf s = \dfrac{25}{5} }$

☞ $\blue {\sf s = 5 }$

Therefore, Distance Travelled is 5 m

Therefore

$\purple{\sf \dag Acceleration = 2.5 m \: s^{-2} \dag }$

$\purple {\sf \dag Distance \: Travelled = 5 \: m \dag }$

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Answer 2

Given :

• Initial velocity (u) = 0
• Final velocity (v) = 5m/s
• Time (t) = 2s

To find :

• Acceleration (a)
• Distance (s)

Solution :

✞According to the first equation of motion

$\bigstar{\large{\boxed{\bf{\red{v=u+at}}}}} \\ \\ \implies\large\tt 5=0+a\times{2} \\ \\ \\ \implies\large\tt 5=2a \\ \\ \\ \implies\large\tt a=\cancel\dfrac{5}{2} \\ \\ \\ \implies\large\tt a=2.5m/s^2$

Hence,

• Acceleration of car is 2.5m/s²

$\rule{200}3$

✞Now, according to the second equation of motion

$\bigstar{\large{\boxed{\bf{\red{s=ut+\dfrac{1}{2}at^2}}}}} \\ \\ \implies\large\tt s=0\times{2}+\dfrac{1}{2}\times{2.5}\times{(2)^2} \\ \\ \\ \implies\large\tt s=0+\dfrac{1}{2}\times{2.5}\times{4} \\ \\ \\ \implies\large\tt s=2.5\times{2} \\ \\ \\ \implies\large\tt s=5m$

Hence,

• Distance covered by a car is 5m

$\rule{200}3$