Question

The acceleration-time graph of a particle moving in a straight line is as shown in figure. The velocity of the particle at time t = 0 is 2 m/s. The velocity after 2 seconds will be ?

()

options :-

(1) 6 m/s
(2) 4 m/s
(3) 2 m/s
(4) 8 m/s

please answer with full explanation

note: all lines in the graph are straight

Answer 1

I got
(1)

Explanation:

From Graph: `dota=4\ ms^-3` from `t=0` to `t=1\ s`
and `dota=-4\ ms^-3` from `t=1` to `t=2\ s`

1. from `t=0` to `t=1\ s`
   `dota=4\ ms^-3`
   Integrating both sides with time we get

   `int\ dota\ dt=int\ 4\ dt`
   `=>a=4t+C`
   where `C` is a constant of integration to be calculated from initial conditions.
   From Graph at `t=0`, `a=0`
   `:.a=4t\ ms^-2` ........(1)
   Rewriting this as
   `a=dotv=4t\ ms^-2`
   Integrating both sides with time we get
   `int\ dotv\ dt=int\ 4t\ dt`
   `=>v=4t^2/2+C_1`
   `=>v=2t^2+C_1`
   where `C_1` is a constant of integration to be calculated from initial conditions. Given, at `t=0` `v=2\ ms^-1`. Inserting these values in above expression we get
   `=>2=2(0)^2+C_1`
   `=>C_1=2`
   This gives us the expression for velocity as
   `v=2+2t^2` .......(2)
   and `v_(t=1)=4\ ms^-1` ...........(3)

2. from `t=1` to `t=2\ s`
   `dota=-4\ ms^-3`
   Integrating both sides with time we get

   `int\ dota\ dt=int\ -4\ dt`
   `=>a=-4t+C_2`
   where `C_2` is a constant of integration to be calculated from initial conditions at `t=1\ s`.
   From Graph at `t=1`, `a=4\ ms^-2`. Inserting in above expression we get
   `4=-4xx1+C_2`
   `=>C_2=8`
   `:.a=-4t+8\ ms^-2` ........(4)
   Rewriting this as
   `a=dotv=-4t+8\ ms^-2`
   Integrating both sides with time we get
   `int\ dotv\ dt=int\ (-4t+8)\ dt`
   `=>v=-4t^2/2+8t+C_3`
   `=>v=-2t^2+8t+C_3`
   where `C_3` is a constant of integration to be calculated from calculated initial conditions at `t=1` `v=4\ ms^-1`. Inserting these values in above expression we get
   `=>4=-2(1)^2+8xx1+C_3`
   `=>C_3=-2`
   This gives us the expression for velocity as
   `v=-2t^2+8t-2` .......(3)
   and `v_(t=2)=(-2(2)^2+8xx2-2)`
   `=>v_(t=2)=6\ ms^-1` ...........(6)