Question

2. The velocity-time (x-1) graph for a particle moving along x-axis is shown in the figure. The acceleration at
t=4.5 s is
(1) 2 m/s2
(2) 4 m/s2
(3)8m/s2
(4)6m/s2
plzz solve this question with explanation​

Answer 1

Answer:

ANS IS 4m/s²

Explanation:

1) consider the down wala triangle as

ABC TRAINGLE -

1/2 BASE × HT.= 1/2×4×(-16) = -32 .

then consider up wala traingle.

1/2 ×12×16= 96

therefore

acc.= 64/ 16 = 4m/s².

Answer 2

Answer:

The value of the acceleration at $t=4.5s$ will be equal to $4ms^{-2}$.

Therefore the option (2) is correct.

Explanation:

We know $Acceleration,a=\frac{dv}{dt}$

In the velocity time graph (attached below),

We can see that $line AB$ represent the the change in velocity w.r.t. time,

At $t=4.5sec$

$Acceleration,(a)=\frac{dv}{dt}$ = slope of AB

$a=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}$

$a=\frac{16-0}{4-0} =\frac{16}{4}}$

$a=4ms^{-2}$

Therefore the acceleration of particle at $t=4.5sec$ is equal to $4ms^{-2}$.