Question

11. Study the velocity-time graph given below and answer the questions based on it.

picture down

(pls help me)

Answer 1

Answer:

Acceleration for 2 to 4 sec.=-2 m/sec^2

Total displacement = 12 metre

Explanation:

Since we know that the $acceleration(a)=\frac{dv}{dt}$

for first 2 sec velocity is constant so acceleration for first two second is zero.

so, we are calculating the acceleration for 2 to 4 sec.

$\frac{dv}{dt}\ =\ \frac{0-4}{4-2}\ =\ \frac{-4}{2}\ =\ -2$

Acceleration for 2 to 4 sec.=-2 m/sec^2

since we know that,

v x t = S

for first two second

S = 4 x 2 = 8 metre.

for next two second we are calculating the area of triangle for time t = 2 to t = 4 sec.

The area of triangle  =$\frac{1}{2}\times base\times height$

                            =$\frac{1}{2}\times 2 \times 4$

                            = 4 metre

So, total displacement = 12 metre.