Question

A car starts from rest and accelerates uniformly. If it travels a distance of 4m in the first second, what distance will it travel in the 2nd second and the 3rd second?

Answer 1

Solution :

⏭ Given:

✏ Initial velocity of car = 0

✏ Distance covered by carin 1st second = 4m

⏭ To Find:

✏ Distance covered by car in 2nd and 3rd second.

⏭ Formula:

✏ Formula of distance covered by particle in nth second is given by

$\star \: \underline{ \boxed{ \tt{ \red{d_ n = u + \dfrac{a}{2} (2n - 1)}}}} \: \star$

⏭ Terms indication:

• u denotes initial velocity
• a denotes acceleration
• n denotes no. of second
• $\tt{d_n}$ denotes distance covered in nth second

⏭ Calculation:

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$\mapsto \sf \: 4 = 0 + \dfrac{a}{2} {\large{[}}2(1) - 1{ \large{]}} \\ \\ \mapsto \sf \: 4 = \dfrac{a}{2} (2 - 1) \\ \\ \mapsto \sf \: 4 = \dfrac{a}{2} (1) \\ \\ \mapsto \: \boxed{ \tt{ \blue{a = 8 \: m {s}^{ - 2}}}}$

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• Distance covered in 2nd second

$\mapsto \sf \: d_2 = 0 + \dfrac{8}{2} [2(2) - 1]\\ \\ \mapsto \sf \: d_2 = 4(4 - 1) \\ \\ \mapsto \sf \: d_2 = 4(3) \\ \\ \mapsto \: \boxed{ \tt{ \green{d_2 = 12 \: m}}}$

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• Distance covered in 3rd second

$\mapsto \sf \: d_3 = 0 + \dfrac{8}{2} [2(3) - 1] \\ \\ \mapsto \sf \: d_3 = 4(6 - 1) \\ \\ \mapsto \sf \: d_3 = 4(5) \\ \\ \mapsto \: \boxed{ \tt{ \orange{d_3 = 20 \: m}}}$

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Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{s_{2}=12\:m}}}$

$\green{\tt{\therefore{s_{3}=20\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given: }} \\ \tt: \implies Initial \:velocity(u) = 0 \: m/s \\ \\ \tt: \implies Distance \: travel \: in \: 1st \: sec = 4 \: m \\ \\ \red{\underline \bold{To \: Find: }}\\ \tt: \implies Distance \: travel \: in \: 2nd \: sec = ? \\ \\ \tt: \implies Distance \: travel \: in \:3rd\: sec = ?$

• According to given question :

$\tt \circ \: Initial \: velocity = 0 \: m/s \\ \\ \tt \circ \: n = 1 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{n} = u + \frac{a}{2} (2n - 1) \\ \\ \tt: \implies s_{1} =0 + \frac{a}{2} (2 \times 1 - 1) \\ \\ \tt: \implies 4 = \frac{a}{2} \\ \\ \green{\tt: \implies a = 8 \: {m/s}^{2} } \\ \\ \bold{For \: 2nd \: second : } \\ \tt: \implies s_{2} =0 + \frac{8}{2} (2 \times 2 - 1) \\ \\ \tt: \implies s_{2} =4 \times 3 \\ \\ \green{\tt: \implies s_{2} =12 \: m} \\ \\ \bold{For \: 3rd\: second : } \\ \tt: \implies s_{3} =0 + \frac{8}{2} (2 \times 3 - 1) \\ \\ \tt: \implies s_{3} =4 \times 5 \\ \\ \green{\tt: \implies s_{3} = 20\: m}$