A moving body is covering distances which are proportional to square of the time. Then the acceleration of…
A moving body is covering distances which are proportional to square of the time. Then the acceleration of the body is
- decreasing.
- Constant but not Zero
- Zero
- Increasing
Solution
$s \propto t^{2}$
$s=k t^{2}$
$\frac{d s}{d t}=2 k t$
$\frac{d^{2} s}{d t^{2}}=2 k$
Asked in: MHT CET 2020 (16 Oct Shift 1)
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