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d c pandey objective physics...d c pandey objective physics...

2 Answers
Sep 29, 2017

Velocity v(ms^-1) satisfies 3.16<= v<= 3.78 and b) is the best answer.

Explanation:

Calculating the upper and lower bound helps you in this type of problem.

If the body travels the longest distance (14.0 m) in the shortest
time (3.7 s), the velocity is maximized. This is the upper bound
of the velocity v_max

v_max = (14.0 (m))/(3.7 (s)) = 3.78(ms^-1).

Simirally, the lower bound of the velocity v_min is obtained as
v_min = (13.6 (m))/(4.3 (s)) = 3.16(ms^-1).

Therefore, the velocity v stands between 3.16(ms^-1) and 3.78(ms^-1). Choice b) fits this best.

Sep 29, 2017

Option (b)
(3.45+-0.30)m/s

Explanation:

if the Quantity is defined as x=a/b
let Deltaa= "Absolute error for a"
Deltab= "Absolute error for b"
Deltax= "Absolute error for x"
then The Maximum Possible Relative Error in x is
(Deltax)/x=+-[(Deltaa)/a+(Deltab)/b]

Now
Distance =(13.8+-0.2)m
s=13.8 m and Delta s=0.2m
Time =(4.0+-0.3)m
t=4.0 m and Delta t=0.3m

Velocity of Body Within error Limit is v+Deltav
Now "velocity"="Distance"/"time"
v=s/t=13.8/4=3.45 m/s

and Relative error in Velocity
(Deltav)/v=+-[(Deltas)/s+(Deltat)/t]
(Deltav)/v=+-[(0.2)/13.8+(0.3)/4]=0.014+0.075=0.089
Absolute Error in Velocity
Deltav=0.089xxv=0.089xx3.45=0.307 m/s

Hence

Velocity of Body Within error Limit is
v+Deltav=(3.45+-0.30)m/s

Option (b)

Hope You gets your answer.