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- From the velocity-time graph of uniformly accelerated motion, establish the equation $ \mathrm{S}=\mathrm{ut}+1 / 2 \mathrm{at}^{2} $.
- A bus starting from rest moves with a uniform acceleration of $ 0.2 \mathrm{~m} / \mathrm{s}^{2} $ for 2 minutes. Find
- The speed acquired
- The distance travelled
- The speed acquired
- The distance travelled
(a)
Derivation of the second equation of motion: $s=ut+\frac{1}{2}\times a{t}^{2}$ by graphical method -
Let's assume that the distance travelled 's' by the body in time 't' can be calculated by calculating the area under the velocity-time graph.
The area under the graph is equal to the area of OABC.
Thus,
Distance travelled = area of figure OABC
$=Area\ of\ triangle\ OADC+Area\ of\ triangle\ ABD$
$=(OA\times OC)+\left(\frac{1}{2}\times AD\times BD\right)$ $\left(\because Area\ of\ rectangle=length\times breadth\ and\ Area\ of\ triangle=\frac{1}{2}\times base\times height\right)$
$=(u\times t)+\left(\frac{1}{2}\times t\times at\right)$ $=ut+\frac{1}{2}a{t}^{2}$
So, distance travelled $s=ut+\frac{1}{2}a{t}^{2}$
$s=ut+\frac{1}{2}a{t}^{2}$
Hence, the second equation of motion is derived by the graphical representation.
b) Given,
Acceleration, a = 0.2m/s2
Time, t = 2 minutes = 2 x 60 = 120 sec [converted minute into second]
Initial velocity = u = 0m/s [Since bus starts from rest]
Solution
(i) Speed Acquired
Speed acquired will be the final velocity, which can be given as-
Speed acquired = Final velocity = v
Since we have u, a and t, then we can apply first equation of motion,
$v=u+at$
$v=0+0.2\times 120$
$v=24m/s$
Thus, speed acquired will be 24m/s.
(ii) Distance travelled
We find Distance travelled by 2nd equation of motion
$s=ut+\frac{1}{2}a{t}^{2}$
substituting the given values we get-
$s=0\times 120+\frac{1}{2}\times 0.2\times (120{)}^{2}$
$s=0+\frac{1}{2}\times \frac{2}{10}\times 14400$
$s=1440$
Thus, distance travelled will be 1440m.
Updated on: 10-Oct-2022
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