Question

please answer this question fastly 16th question please​

Answer 1

Solution :

⏭ Given:

✏ A car starts from rest and moving along x-axis with constant acceleration 5m$\sf{s^{-2}}$ for 8 seconds.

✏ It then continues with constant velocity.

⏭ To Find:

✏ Total distance covered by car in 12 seconds since it started from the rest.

⏭ Formula:

✏ Formula of distance covered in acceleratory motion is given by

$\bigstar \: \boxed{ \tt{ \pink{d = ut + \frac{1}{2} a {t}^{2} }}} \: \bigstar$

✏ Formula of distance covered uniform motion is given by

$\bigstar \: \boxed{ \tt{ \purple{d = v \times t}}} \: \bigstar$

⏭ Calculation:

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$\circ \sf \: d_1 =( 0 \times 8) + { \huge{(}} \dfrac{1}{2} \times 5 \times {8}^{2} { \huge{)}} \\ \\ \circ \sf \: d_1 = 0.5 \times 5 \times 64 \\ \\ \circ \: \boxed{ \tt{ \red{ \large{d_1 = 160 \: m}}}}$

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$\circ \sf \: d_2 = v \times t \\ \\ \circ \sf \: d_2 = (u + at) \times t' \\ \\ \circ \sf \: d_2 = (0 + 40) \times 4 \\ \\ \circ \: \boxed{ \tt{ \blue{ \large{ d_2 = 160 \: m}}}}$

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$\implies \sf \: d_t = d_1 + d_2 \\ \\ \implies \sf \: d_t = 160 + 160 \\ \\ \implies \: \boxed{ \tt{ \large{ \red{d_t = 320 \: m}}}} \: \orange{ \bigstar}$

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Answer 2

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

$\green{\tt{\therefore{Total\:distance\:travelled=320\:m}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Acceleration = 5 \: {m/s}^{2} \\ \\ \tt: \implies Time(t) = 8 \: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Distance \: travel \: in \: 12 \: second = ?$

• According to given question :

$\tt \circ \: Initial \: velocity = 0 \: m/s \\ \\ \bold{as \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ \tt: \implies s_{1} = 0 \times 8 + \frac{1}{2} \times 5 \times {8}^{2} \\ \\ \tt: \implies s_{1} = 5 \times 32 \\ \\ \green{\tt: \implies s_{1} =160 \: m} \\ \\ \bold{For \: final \: velocity : } \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {v}^{2} = {0}^{2} + 2 \times 5 \times 160 \\ \\ \tt: \implies {v}^{2} = 1600 \\ \\ \tt: \implies v = \sqrt{1600} \\ \\ \green{\tt: \implies v = 40 \: m/s} \\ \\ \tt \circ \: New \: Initial \: velocity = 40 \: m/s \\ \\ \tt \circ \: Acceleration = 0 \: m/ {s}^{2} \\ \\ \tt \circ \:Time = 4 \: sec \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ \tt: \implies s_{2} = 40 \times 4 \\ \\ \green{\tt: \implies s_{2} =160 \: m} \\ \\ \bold{For \: total \: distance : } \\ \tt: \implies s= s_{1} + s_{2} \\ \\ \tt: \implies s= 160 + 160 \\ \\ \green{\tt: \implies s= 320 \: m} \\ \\ \green{\tt \therefore Total \: distance \: travelled \: by \: car \:is \: 320 \: m}$