Question

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Answer 1

$\huge{\sf{\red{\underline{Question:-}}}}$

A car starts from rest and coasts down a hill with a constant acceleration. It goes 90m in 8s.

$\huge{\sf{\blue{\underline{To\:find:-}}}}$

• Acceleration
• velocity after 8s

$\huge{\sf{\orange{\underline{Formula\:used:-}}}}$

• s = ut + $\frac{1}{2}$at²
• v = u + at

$\huge{\sf{\green{\underline{Answer:-}}}}$

$\sf{Let \:the \:initial \:velocity\: be \:u \:= \:0m/s}$

$\sf{Let \:the \:distance\: covered\: be \:s \:= \:90m}$

$\sf{Let\: the\: time \:taken\: be \:t\: =\: 8s}$

$\sf{Let\: the\: acceleration \:be \:a}$

$\sf{Let\: the\: final\: velocity\: be\: v}$

$\underline{\bf{Finding \:a:}}$

s = ut + $\frac{1}{2}$at²

$⟶\:\sf{90 \:= \:0\: *\: 8 \:\: + \:\: \frac{1}{2}a * 8 * 8}$

$⟶\:$90 = 0 + 32a

$⟶\:$90 = 32a

$⟶\:$a = $\bcancel{\frac{90}{32}}$

$⟶\:$a = 2.8125 m/s²

$\underline{\bf{Finding \:v:}}$

v = u + at

$⟶\:$v = 0 + 2.8125 * 8

$⟶\:$v = 2.8125 * 8

$⟶\:$v = 22.5 m/s

Answer 2

answer :

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