Question

A roller coaster is traveling at 13 m/s when it approaches a hill that is 400 m long. Heading down the hill, it accelerates at 4.0 m/s2. What is the final velocity of the roller coaster?.

Answer 1

u=13m/s
h=400m
a=4m/s^2
v=?
u can apply Newton s 3rd law of motion
v^2=169+3200

Answer 2

Aɴꜱᴡᴇʀ

☞ The final velocity of the roller coaster is 58.04 m/s

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Gɪᴠᴇɴ

➳ Initial velocity (u) = 13 m/s

➳ Distance (s) = 400 m

➳ Acceleration = 4 m/s²

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Tᴏ ꜰɪɴᴅ

➤ The final velocity of the body?

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Sᴛᴇᴘꜱ

The acceleration of the body can be found by the third equation of motion, that is,

$\large {\underline{ \boxed{ \sf {v}^{2} - {u}^{2} = 2s}}}$

So, substituting the given values,

$\large \tt \leadsto{} {v}^{2} = {13}^{2} + 2 \times 4 \times 400 \\ \\ \large \tt \leadsto {v}^{2} = 169 + 3200 \\ \\ \large \tt \leadsto {v}^{2} = 3369 \\ \\ \large \tt \leadsto{}v = \sqrt{3369} \\ \\ \large \tt { \pink{ \leadsto{}v = 58.04 \: m/s}}$

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