Question

A particle starts from rest with acceleration 2m/s^2. The acceleration of the particle decreases down to zero uniformly during time interval of 4s. The velocity of particle after 2s is?

Answer 1

Here we will apply the equation of motion v=u+at
U is 0 a is given and time is 2 sec.
Therefore,v=0+2×2
=4m/s is the velocity

Answer 2

Answer:

3 m/s

Explanation:

a(t) = mt + c

a(0) = 2   ==> m(0) + c = 2

              ==> c = 2

also given a(4) = 0   ==> 4m + 2 = 0

                                 ==> m = $\frac{-1}{2}$

so a(t) = -1/2m + 2

integrating ==> $\int\limits^t_T {\frac{dv}{dt} }$ = a

                 ==> $\int\limits^v_V  dv$  =  $\int\limits^t_T {a.t} \, dt$

                 ==> v - V = $\frac{-t^{2}- T^{2}  }{4}$ + 2(t-T)

                 ==> v = $\frac{-t^{2} }{4}$ + 2t .................. (As V = 0 and T = 0)

                 ==> v = $\frac{-2^{2} }{4}$ + (2 × 2)

                 ==> v = -1 + 4

                 ==> v = 3 m/s