Question

A bus starting from rest moves qith a uniform acceleration of 0.1 m s^-2 for 2 minutes.
Find :
A ) The speed acquired
B ) The distance travelled

Answer 1

$\Large \tt \mid \mid{solution} \mid \mid$

( A ) The bus starts from the rest .

$\tt \therefore \: initial \: velocity \: (u) \: = \: 0 \: \frac{m}{s}$

$\tt \: acceleation \: ( \: a \: ) \: = 0.1 \: m.s {}^{ - 2}$

$\tt \: time \: = \: 2 \: minutes \: = 120 \: seconds$

$\tt \: accelration \: is \: given \: by \: \\ \tt the \: equation \: \large \tt{a \: = \frac{v - u}{t} }$

$\tt \therefore \: terminal \: velocity \: \large \tt{ ( \: v \: ) \: = (at) \: + \: u}$

$\tt= \:(0.1 \times 120) + 0$

$\tt = 12 + 0$

$\tt \therefore \: terminal \: velocity \: (v) \: = \: 12 \: seconds$

( B ) As per the third motion equation

$\tt \: since \: a \: = \: 0.1 \: ms{}^{ - 2}, \: v \: = m.s {}^{ - 1} , \: u \: = \: 0 \: m.s {}^{ - 1}, \: \\ \tt \: the \: following \: value \: for \: ( \: distance \: ) \: can \: be \: obtained.$

$\tt \: distance \: , \large \tt s \: = \: \frac{v {}^{2} - u {}^{2} }{2a} \\ \large \tt \: = \frac{12 {}^{2} - 0 {}^{2} }{2 \times 0.1}$

$\tt \therefore \: s \: = 720 \: m \:$

$\tt \: The \: speed \: acquired \: is \: 12 \: m.s {}^{ - 1} \: and \: the \: total \: distance \: travelled \: is \: 720 \: m$

Note 1 : in the Above answer A = Acceleration , U = Initial velocity , T = Time and S = Distance

Note 2 : Swipe left side to see the full answer

Answer 2

Answer:

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