1.

A particle moving in a straight line with uniform acceleration has a velocity 10 cm/sec and 8 seconds later has a velocity 54 cm/sec. What will be the value of space described?(a) 172 cm(b) 176 cm(c) 178 cm(d) 174 cmI have been asked this question in semester exam.My doubt stems from Calculus Application topic in portion Application of Calculus of Mathematics – Class 12

Answer»

The correct answer is (b) 176 cm

For EXPLANATION I would say: LET, the particle moving with a uniform acceleration of f cm/sec^2.

By question initial velocity of the particle is u = 10 cm/sec and its velocity after 8 SECONDS = v = 34 cm/sec.

Thus, using the formula v = u + ft we get

34 = 10 + f*8

Or 8f = 24

Or f = 3

Therefore, the required acceleration of the particle is 3 cm/sec^2.

Thus, the space described by the particle in 8 seconds,

= [10*8 + 1/2(3)(8*8)[using the formula s = ut +1/2(ft^2)]

= 80 + 96

= 176 cm.



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