Question

The velocity (V) versus position (x) graph of a particle moving in a straight line is
shown. The acceleration of the particle at x = 1 m is -p m/s? Find the value of p.
सरल रेखा में गतिशील एक कण का वेग (v) तथा स्थिति (x) आरेख दर्शाया गया है। x 1m पर कण का
त्वरण-pm/s' है।p का मान ज्ञात कीजिए।
v(m/s)+
+x (m)​

Answer 1

Answer:

average velocity is 25m/s so distance will be 25×20=500m

now finding area of the graph between velocity and time graph we get

area of two triangles + area of rectangle.

2. 21

t.5.t+(25−2t).5t=500=>t ^2−25t+100

=>t=5

so option B is answer.