The equation for the change in the position of a train (measured in units of length) is given by the following expression: x = ½ at² + bt³ where a and b are constants and t is time (measured in units of time). if (mass), (length), and (time) represent units of mass, length and time respectively then b must have units of:
(a) (length)/(time)
(b) (length)/(time³)
(c) (mass)(length)
(d) (length)/(mass)
(e) (mass2)(time2)
(f) 1/(mass)(time)
(g) 1/(length)(time)
(h) (time)(mass)
(i) (mass)(time2)
(j) 1/(time³)
(k) (time)/(mass)
(l) (length)/(time²)
(mass)(length2)

Explanation:

The position of a train is given as a function of time

x(t) = ½at² + bt³

Where a and b are constant

We want to know the dimensional unit of b.

x is length and has a unit of metre

t is time and has a unit of seconds

b is a constant with unknown unit

a is a constant with unknown unit

Given that,

x = ½at² + bt³

x(metre) = ½ a t²(seconds)² + bt³(seconds)³

For x to be dimensionally correct, a and b must have a unit that cancel out the time

So, a•t², so cancel out the time square "a" must have a unit that cancel out the time square. Also a must have "a" unit of distance too. So, "a" will have m/s² unit

Then, at² will give

a (metre/seconds)² × t² (seconds)²

Also, for "b", same procedure but b wants to cancel out seconds cube, so instead of seconds square division, " b" will have secondds cube division I.e m / s³

Then, bt³ will give

b(metre/seconds³) × time (seconds)³

So, the unit of b is metre/seconds³

Since Length = metre

And Time = seconds

Which is Length/ Time³

So, the correct answer is B.

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