Question

Can anyone please do the 29th one please

Answer 1

Answer:

Case 1

Given that,

v = 10m/s

u = 0

t = 25s

By 1st equation of motion,

v = u + at

Substituting the values of v , u and t,

10 = 0 + 25a

25a = 10

a = 10/25

a = 0.4m/s²

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Case 2

Given that,

v = 0

u = 10m/s

t = 50s

By 1st equation of motion,

v = u + at

Substituting the values of v, u and t,

0 = 10 + 50a

50a = -10

a = -0.2m/s²

Hope it helps

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Answer 2

$\rm{ \huge Given :( \bf \: Case_{ \small1} )}$

• u = 0 m/s
• v = 10 m/s
• t = 25 second

Therefore,

$\huge \sf{a = \frac{v - u}{t} } \\ \\ \boxed{ \frak{= > \frac{(10 - 0)}{25} = \frac{10}{25} = \purple{0.4} \: {ms}^{ - 2} }}$

Now,

$\rm \huge Given :({ \bf{ Case_{ \small2}})}$

• u = 10 m/s
• v = 0 m/s
• t = 50 second

$\huge \sf{a = \frac{v - u}{t} } \\ \\ \tt{ = > \frac{0 - 10}{50} = \frac{ - 1 \cancel0}{ 5 \cancel0} = \frac{ - 1}{5} \: \: {ms}^{ - 2} } \\ \\ \boxed{\frak{ = > \purple{ - 0.2} \: \: {ms}^{ - 2} }}$

Then,

$\sf\huge{ S_{1} = ut + \frac{1}{2} {at}^{2} } \\ \\ = > 0(25) + \frac{1}{2} \times 0.4 \times {(25)}^{2} \\ \\ = > 0 + \frac{ \cancel{0.4 }\times 25 \times 25}{ \cancel2} \\ \\ = > 0.2 \times 625 = \frak{\red{125} \: m}$

And,

$\huge \sf{S_{2} = ut + \frac{1}{2} {at}^{2} } \\ \\ = > 10(50) + \frac{1}{2} \times ( - 0.2) \times {(50)}^{2} \\ \\ = > 500 + \frac{ \cancel{-0.2} \times 2500}{ \cancel2} \\ \\ = > 500 + ( - 0.1 \times 2500) \\ = > 500 + ( - 250) \\ = > 500 - 250 = \frak{ \red{250} \: m}$

Hence,

$\rm{Total \: distance = (250 + 125) \: m} \\ \\ \boxed{\huge = > \: \: \: \: \: \frak{\red{375} \: m}}$