Question

A car starts from the rest and accelerates at the rate of 5m/s2. It continue
accelerating for 10 s and then move at constant velocity for next 10 s. Calculate
the total distance traveled by the car in 20s.

Answer 1

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

$\green{\tt{\therefore{Total\:Distance=750\:m}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Acceleration(a) = 5 { \: ms}^{2} \\ \\ \tt: \implies Time(t_{1}) = 10 \: sec \\ \\ \tt: \implies Time(t_{2}) = 10 \: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Distance \: travelled(s) = ?$

• According to given question :

$\tt \circ \: Initial \: velocity = 0 \: m/s \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ \tt: \implies s_{1}= 0 \times 10 + \frac{1}{2} \times 5 \times {10}^{2} \\ \\ \green{\tt: \implies s_{1} = 250 \: m} \\ \\ \bold{Velocity \: after \: 10 \: sec} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {v}^{2} = {0}^{2} + 2 \times 5 \times 250 \\ \\ \tt: \implies {v}^{2} =2500 \\ \\ \tt: \implies v = \sqrt{2500} \\ \\ \green{\tt: \implies v =50 \: m/s} \\ \\ \tt \circ \: Initial \: velocity \: after \: 10 \: sec = 50 \: m/s \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s =ut + \frac{1}{2} {at}^{2} \\ \\ \tt: \implies s_{2} =50 \times 10 + \frac{1}{2} \times 0 \times {10}^{2} \\ \\ \green{\tt: \implies s_{2} =500 \: m} \\ \\ \bold{For \: Total \: Distance : } \\ \tt: \implies s = s_{1} + s_{2} \\ \\ \tt: \implies s =250 + 500 \\ \\ \green{\tt: \implies s =750 \: m}$

Answer 2

Aɴꜱᴡᴇʀ

➜$\rm{\orange{S = 750 \:m}}$

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Gɪᴠᴇɴ

$\sf \bigstar{} \: \: initial \: velocity = 0m/s \\ \\ \sf \bigstar \: \: {}acceleration = 5m / {s}^{2} \\ \\ \sf \bigstar{} \: \: t_1 =10 \: seconds \\ \\ \sf \bigstar{} \: \: t_2 = seconds$

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ᴛᴏ ꜰɪɴᴅ

We have to find the distance travelled by the body

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Sᴛᴇᴘꜱ

As it has been give on that u = 0 we can first use the formula $\tt{s = ut+\frac{1}{2} a{t}^{2}}$

$\tt \leadsto{}s_1 =0 \times 10 + \frac{1}{2} \times 5 \times {(10)}^{2} \\ \\ \tt \leadsto{}s_1 = \frac{1}{ \cancel2} \times 5 \times \cancel{100} \\ \\ \tt \leadsto{ \red{s_1 =250 \: \sf{}m }}$

Now lets find the velocity of the body after 10 seconds

So now let's use the formula $\sf{v}^{2}-{u}^{2}=2as$

$\tt{}it \: can \: also \: be \: written \: as \: \bf {v}^{2} = {u}^{2} + 2as \\ \\ \tt{} \hookrightarrow{} {v}^{2} = {0}^{2} + 2(10)(250) \\ \\ \tt \hookrightarrow{} {v}^{2} = 2500 \\ \\ \tt{} \hookrightarrow{}v = \sqrt{2500} \\ \\ \tt{} \hookrightarrow{ \blue{v = 50}}$

Actually the value that we found now is the initial velocity after 10 seconds

So now lets find the 2nd distance

$\tt{}lets \: use \: the \: formula \: \sf{}s = ut + \frac{1}{2} a {t}^{2} \\ \\ \tt{} \implies{}s_2 = 50 \times 10 + \frac{1}{2} (0)(10 {)}^{2} \\ \\ \tt{} \implies{} \green{s_2 = 500 \sf{} \: m}$

So now the total distance covered is equal to,

$\tt{}s_1 + s_2 = s \\ \\ \tt \dashrightarrow{}250 + 500 \\ \\ \tt{} \purple{ \dashrightarrow{}750 \: \sf{}m}$

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