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Question

Physics Question on Motion in a straight line

Acceleration of a particle is given by an equation, a=2+2t+3t2a = 2 + 2t + 3t^2 when tt is the time taken. If the particle starts at t=0t = 0 with velocity 2ms1 2 \,ms^{-1}, then the velocity in ms1m s^{-1} at the end of 2s2 s is

A

12

B

18

C

27

D

36

Answer

18

Explanation

Solution

Here a = 2 + 2t + 3t23t^2 i.e. dυdt=2+2t+3t2\frac{d \upsilon}{dt} = 2 + 2t + 3t^2 i.e. uυdυ=0t(2+2t+3t2)dt\int\limits^{\upsilon}_{u} \, d \upsilon = \int\limits^t_0 \, (2 + 2t+ 3t^2) dt υu=(2+2t+3t2)dt=2t+2t22+3t33=2t+t2+t3\upsilon - u = \int\limits \, (2 + 2t + 3t^2 )dt = 2t +\frac{2t^2}{2} + \frac{3t^3}{3} = 2t + t^2 + t^3 = υ=2t+t2+t3+u=2×2+22+23+2=18ms1\upsilon =2t + t^2 + t^3 + u = 2 \times 2 + 2^2 + 2^3+2 = 18 \, m \, s^{-1}