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)

(B) (length) / (time³)

Explanation

The equation x = ½ at² + bt³ has to be dimensionally correct. In other words the term bt³ and ½ at² must have units of change of position = length.

We solve in order to find the dimension of b:

[x]=[b]*[t]³

length=[b]*time³

[b]=length/time³

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.