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Standard XII

Physics

Distance

A body starti...

Question

A body starting from rest moves with an acceleration of 2 $m / s^{2}$ for the first 5 seconds and increases its acceleration to 4 $m / s^{2}$ for the next 8 seconds. Calculate the total distance covered in metres.

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Solution

Let $S_{1}$ be the distance covered in first 5 seconds.

$S_{1}$ = ut + $\frac{{a t}^{2}}{2}$

= (0 x 5) + $\frac{2 x 5^{2}}{2}$

= 25m

Let $S_{2}$ be the distance covered in next 8 seconds.

Initial velocity = 0 + (2 x 5) m/s

= 10 m/s

a = 4 $m / s^{2}$ , t = 8s

$S_{2}$ = (10 x 8) + $\frac{4 x 8^{2}}{2}$

= 208m

Total Distance = $S_{1} + S_{2}$

= (25 + 208)m

= 233m

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A body starting from rest moves with an acceleration of 2 m/s2 for the first 5 seconds and increases its acceleration to 4 m/s2 for the next 8 seconds. Calculate the total distance covered in metres. __

Let S1 be the distance covered in first 5 seconds. S1 = ut + at22 = (0 x 5) + 2 x 522 = 25m Let S2 be the distance covered in next 8 seconds. Initial velocity = 0 + (2 x 5) m/s = 10 m/s a = 4 m/s2 , t = 8s S2 = (10 x 8) + 4 x 822 = 208m Total Distance = S1+S2 = (25 + 208)m = 233m

A body starting from rest moves with an acceleration of 2 $m / s^{2}$ for the first 5 seconds and increases its acceleration to 4 $m / s^{2}$ for the next 8 seconds. Calculate the total distance covered in metres.

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