Given, the initial velocity (u) = 80km/hour = \( \frac{80,000m}{3600s}=22.22\,m.s^{-1}\)
The final velocity (v) = 60km/hour \(= \frac{60,000m}{3600s}=16.66\,m.s^{-1}\)
Time frame, t = 5 seconds.
Therefore, acceleration (a) \( =\frac{v-u}{t}\) \( =\frac{16.66\,ms^{-1}-22.22\,m.s^{-1}}{5s}\)
\( =-1.112\,m.s^{-2}\)
Therefore, the total acceleration of the bus is \(-1.112\,m.s^{-2}\). It can be noted that the negative sign indicates that the velocity of the bus is decreasing.
Answered by Vishal kumar | 2 years agoA bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find
(a) the speed acquired,
(b) the distance traveled.
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What is the quantity which is measured by the area occupied below the velocity-time graph?
A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find
(a) the speed acquired,
(b) the distance travelled.
What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?