an object starting from rest accelerates in a straight line at a constant rate of 5 m/s2 for 10s. calculate the magnitude of displacement during this time
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an object starting from rest accelerates in a straight line at a constant rate of 5 m/s2 for 10s. calculate the magnitude of displacement during this time
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Answer:
Provided that:
To calculate:
Solution:
Using concept:
Using formula:
[tex]{\small{\underline{\boxed{\sf{\rightarrow s \: = ut \: + \dfrac{1}{2} \: at^2}}}}}[/tex]
Where, s denotes displacement or distance or height, u denotes initial velocity, t denotes time taken and a denotes acceleration.
Required solution:
[tex]:\implies \sf s \: = ut \: + \dfrac{1}{2} \: at^2 \\ \\ :\implies \sf s \: = 0(10) + \dfrac{1}{2} \times 5(10)^{2} \\ \\ :\implies \sf s \: = 0(10) + \dfrac{1}{2} \times 5(100) \\ \\ :\implies \sf s \: = 0(10) + \dfrac{1}{2} \times 500 \\ \\ :\implies \sf s \: = 0 + \dfrac{1}{2} \times 500 \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 500 \\ \\ :\implies \sf s \: = 250 \: m \\ \\ :\implies \sf Displacement \: = 250 \: metres[/tex]
Therefore, 250 m is displacement!
Knowledge booster:
[tex]\begin{gathered}\boxed{\begin{array}{c|cc}\bf Distance&\bf Displacement\\\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}&\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}\\\sf Path \: of \: length \: from \: which &\sf The \: shortest \: distance \: between \\ \sf \: object \: is \: travelling \: called \: distance. &\sf \: the \: initial \: point \: \& \: final \\ &\sf point \: is \: called \: displacement. \\\\\sf It \: is \: scalar \: quantity. &\sf It \: is \: vector \: quantity \\\\\sf It \: is \: positive \: always &\sf It \: can \: be \: \pm \: \& \: 0 \: too \end{array}}\end{gathered}[/tex]
Answer:-
We know
[tex]\boxed{\sf s=ut+\dfrac{1}{2}at^2}[/tex]
[tex]\\ \sf \longmapsto s=0(10)+\dfrac{1}{2}5(10)^2[/tex]
[tex]\\ \sf \longmapsto s=\dfrac{5\times 100}{2}[/tex]
[tex]\\ \sf \longmapsto s=\dfrac{500}{2}[/tex]
[tex]\\ \sf \longmapsto s=250m[/tex]