Q#6 (Past Exam Paper – June 2015 Paper 11 Q9)
A body having uniform acceleration a increases its velocity from u to v in time t.
Which expression would not give a correct value for the body’s displacement during time t?
A ut + ½ at2
B vt – ½ at2
C (v+u)(v–u) / 2a
D (v–u)t / 2
Solution:
Answer: D.
Let the displacement be s.
Since the body is undergoing uniform acceleration, we can use the equations of uniformly accelerated motion.
Consider the equation: s = ut + ½ at2 eqn (1) [A is correct]
We know that for uniform acceleration,
v = u + at giving u = v – at eqn (2)
Replace equation (2) in equation (1).
s = (v – at)t + ½ at2 = vt – at2 + ½ at2 = vt – ½ at2 since – at2 + ½ at2 = – ½ at2
[B is correct]
Consider the other equation: v2 = u2 + 2as
Make s the subject of formula
s = v2 – u2 / 2a = (v+u)(v–u) / 2a
since v2 – u2 = (v+u) (v–u)
[C is correct]
The only choice that remains is D and this is incorrect (we will not be able to show this equation as it is not correct).
Choice D would have been correct if there was a positive sign instead. That is, (v+u)t / 2
For uniform acceleration,
Average speed, <v> = (v + u) / 2
Speed = distance / time
giving distance = <v> × t = (v+u)t / 2
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