Question

Acceleration-time graph of a particle is shown in
the diagram, if particle starts with velocity
u = 2.4 m/s, then its velocity at t = 8 s is
pa (m/s)
0.6
0.4 -
>t(s)
2
4
8
(1) 4.2 m/s
(3) 6.2 m/s
(2) 5.2 m/s
(4) 7.2 m/s​

Answer 1

Given that,

Initial velocity = 2.4 m/s

Time = 8 sec

We know that,

Area under acceleration time graph show the change in velocity.

We need to calculate the change in velocity

Using given graph

$\Delta v=\text{area of square+area of trapezium+area of square}$

Put the value into the formula

$\Delta v=0.6\times2+(0.6+0.4)\times2\times\dfrac{1}{2}+0.4\times4$

$\Delta v=3.8\ m/s$

We need to calculate the final velocity

Using formula of change in velocity

$\Delta v=v_{f}-v_{i}$

Put the value of initial velocity

$v_{f}=3.8+2.4$

$v_{f}=6.2\ m/s$

Hence, The final velocity of particle is 6.2 m/s.

(3) is correct option.

Answer 2

Answer:

6.2

Explanation:

see the attached file