MCQ
The displacement is given by $x = 2{t^2} + t + 5$, the acceleration at $t = 2\;s$ is.........$m/{s^2}$
- ✓$4$
- B$8$
- C$10$
- D$15$
Velocity $ = \frac{{dx}}{{dt}} = 4t + 1$
Acceleration $ = \frac{{{d^2}x}}{{d{t^2}}} = 4$ i.e. independent of time
Hence acceleration $ = 4\;m/{s^2}$
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$(1)$ The relation between $\frac{\Delta f}{f}$ and $\frac{\Delta n}{n}$ remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature.
$(2)$ $\left|\frac{\Delta f }{ f }\right|<\left|\frac{\Delta n }{ n }\right|$
$(3)$ For $n =1.5, \Delta n =10^{-3}$ and $f =20 cm$, the value of $|\Delta f |$ will be $0.02 cm$ (round off to $2^{\text {nd }}$ decimal place)
$(4)$ If $\frac{\Delta n }{ n }<0$ then $\frac{\Delta f }{ f }>0$
