Question

Assuming the acceleration in Example 2 above remains constant, how long will Mahendra take to move 1 cm towards virat ?

Answer 1

s=1cm

a=4.001×10^-7

By Newton's second equation

S = ut+1/2at^2

1 =0×t+1\2×4.001×10^-7×t

1/4.001×10^-7=t

4.997=t

5=t

Therefore Mahendra will take 5 seconds to move 1 cm

Answer 2

Answer:Well let us write the data that is given , assuming it is in S.I. units .

s = 1 c.m. = 10−2m.

a = 5.34⋅10−9m/s2.

You are requested to find t .

I shall assume M. starts to move from rest .

So u = 0.

We know that

s=ut+12at2.

0.01=0+125.34⋅10−9⋅t2

=2.67⋅10−9t2.

Now,

2.67=2+0.67=2+23=83.

Now we have

t2=10−283⋅10−9=38⋅10−2−(−9).

t2=38⋅107

t=308⋅106−−−−−−√=308−−√⋅103sec.

Thus it takes

t=1.9364916⋅103=1936.49sec=32min.16.2sec

for M. to reach V.

Hope that helps !

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Explanation:Well let us write the data that is given , assuming it is in S.I. units .

s = 1 c.m. = 10−2m.

a = 5.34⋅10−9m/s2.

You are requested to find t .

I shall assume M. starts to move from rest .

So u = 0.

We know that

s=ut+12at2.

0.01=0+125.34⋅10−9⋅t2=2.67⋅10−9t2.

2.67=2+0.67=2+23=83.

Now we have

t2=10−283⋅10−9=38⋅10−2−(−9).

t2=38⋅107

t=308⋅106−−−−−−√=308−−√⋅103sec.

Thus it takes

t=1.9364916⋅103=1936.49sec=32min.16.2sec

for M. to reach V.

Hope that helps !