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Today I give you three more multiple choice practice questions with solution:

(1) Tom runs from his school to his home with uniform speed

*v*_{1}and returns to his school with uniform speed*v*_{2}. The*average*speed of his round trip is
(a) (

*v*_{1}+*v*_{2})/2
(b) [(

*v*_{1}^{2}+*v*_{2}^{2})/2]^{1/2}^{}
(c) (

*v*_{1}*v*_{2})^{1/2}^{}
(d)

*v*_{1}*v*_{2}/(*v*_{1 }+*v*_{2})
(e)

*2**v*_{1}*v*_{2}/(*v*_{1 }+*v*_{2})
The

*average*speed is the ratio of the total distance traveled to the total time taken.
If the distance between Tom’s school and his home is

*s*, the total distance for the round trip is 2*s*and the total time for the round trip is (*s/**v*_{1}+*s/**v*_{2}).
Therefore,

*average*speed = 2*s/*(*s/**v*_{1}+*s/**v*_{2}) = 2*v*_{1}*v*_{2}/(*v*_{1 }+*v*_{2})
(2) The displacement ‘y’ of a particle thrown vertically down is given by the equation, y

*= 2**t*+ 5*t*^{2}, where y is in metre and*t*is in second. The*average*velocity during the time interval from 2 s to 2.1 s is
(a) 12.5 ms

^{–1}
(b) 16 ms

^{–1}
(c) 20.5 ms

^{–1}
(d) 22.5 ms

^{–1}
(e) 32 ms

^{–1}
The velocity ‘v’ of the particle at the instant

*t*is given by
v = dy

*/*d*t =*2 + 10*t*
Therefore, velocities at instants 2 s and 2.1 s are respectively 2 + 10×2 = 22 ms

^{–1}and 2 + 10×2.1 = 23 ms^{–1}.
The

*average*velocity v_{average}during during the time interval from 2 s to 2.1 s is given by
v

_{average}= (22 + 23)/2 = 22.5 ms^{–1}
[The above question can be answered without using calculus (as in the case of some of the AP Physics B aspirants) like this:

The displacement y is in the form for uniformly accelerated motion in one dimension: y

*=*v_{0}*t*+ ½ a*t*^{2}where v_{0}is the initial velocity and ‘a’ is the acceleration. Therefore, v_{0 }= 2 ms^{–1}and a = 10 ms^{–2}.
The velocity ‘v’ at the instant

*t*is given by v = v_{0}+ a*t*= 2 + 10*t*].
The following question is meant specifically for AP Physics C aspirants:

(3) At the instant

*t*,*the position ‘x’ of a particle moving along the x-axis is given by x = 12**t*^{2}– 2*t*^{3}where x is in metre and*t*is in second. What will be the position of this particle when it moves with the maximum speed along the*positive*x direction?
(a) 32 m

(b) 36 m

(c) 40 m

(d) 48 m

(e) 52 m

The speed

*v*of the particle is given by*v =*dx/d

*t*= 24

*t*– 6

*t*

^{2}.

When the speed is maximum we have d

*v*/d*t*= 0
Therefore, 24

*– 12**t*= 0 from which*t =*2 s.
The position of the particle at 2 seconds will be 12×(2)

^{2}– 2×(2)^{3}= 48 – 16 = 32 m (as obtained from x = 12*t*^{2}– 2*t*^{3}).
Useful posts in this section can be seen at

**physicsplus.**
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