Physics, asked by tys123, 21 days ago

A bus slows down from 43.5 km/h to 34.5 km/h in a school zone. Determine the magnitude of the vehicle's acceleration if it took 12.5 s to slow down.

Write your answer to two decimal places in m/s2.

Answers

Answered by ItzHannu001
3

Answer:

 \tt  \green{ \large \: Given - }

•initial velocity= 43.5 km/h

(in m/s)= 43.5 × 5/18 = 12.08 m/s (u)

• Final velocity= 34.5 km/h

(in m/s) = 9.58 m/s (v)

Time = 12.5 s (t)

 \tt {\green{ \large{To  \: Find :-}}}

Acceleration (a)

 \tt { \large {\green{Formula:-}}}

 \tt {\red {\boxed{ \large{  \tt{a  =  \frac{v - u}{t} }}}}}

Now substituting the values in above formula

 \tt \large \: a =  \frac{9.58 - 12.08}{12.5}  \\  \tt \large \: a =  \frac{ - 2.5}{12.5}  \\  \tt \large \red { a =  - 8.61}

Negative sign shows Negative a acceleration or Retardation

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