please solve it and help me please
Answers
Ques-1 )
a) From pont A to B, distance covered = 300m, Displacement = 300m, Time taken= 2 min 50 sec=(2×60)+50=170 sec.
Average speed = 300/170 = 1.76 m/s
Average velocity = 300/170 = 1.76 m/s
b) From point A to C, Distance covered = 300+100 = 400m, Displacement=300−100=200m, Time taken = 3min 50sec = (3×60)+50 = 230 sec.
Average Speed = 400/230 = 1.73 m/s
Average velocity = 200/230 = 0.87 m/s
Ques-2)
Initial velocity (u) = 0 m/s, Acceleration (a) = 3 m/s^2, Time for which it travels (t) = 8 sec.
Using Newton's second equation of motion,
d = ut + at²/2
d = 0(8) + 3(8)²/2
d = 3*64/2
d = 96 metres.
Ques-3)
a) It's possible. When an object is thrown upwards, at the maximum height the velocity tends to zero but the constant acceleration due to gravity still acts upon it.
b) It's possible too. When an object is executing a circular motion, it's direction of motion is along the tangent and acceleration (centripetal acceleration) is towards the centre of the path.
Ques-4)
Acceleration (a) = 2 cm/s , time duration (t) = 3 sec , initial velocity (u) = 0m/s
Using the equation,
v = u + at
v = 0 + (2)*3
v = 6 cm/s
Ques-5)
Suppose the distance b/w his home and school is x km.
Case-I : When avg speed is 20km/h:
x = (v)t = 20t => t = x/20
Case-II : When avg speed is 40km/h:
x = (v')t' = 30t' => t' = x/40
Average speed = (total distance)/(time taken) = (x+x)/(t+t') = 2x/(x/20 + x/40) = 80/3 = 26.67km/h