How fast is a shadow?

Let the distance of Bob from the streetlight be $a$ , and the distance from the tip of the shadow to the streetlight be $b$ . From similar triangles, $\dfrac{b}{4}=\dfrac{b-a}{1.8}$ , or $20a=11b$ . We take the derivative with respect to time to get $20\dfrac{\text{d}a}{\text{d}t}=11\dfrac{\text{d}b}{\text{d}t}$ . Since $\dfrac{\text{d}a}{\text{d}t}=88\text{ meters per minute}$ , solving for $\dfrac{\text{d}b}{\text{d}t}$ gives $\dfrac{\text{d}b}{\text{d}t}=\boxed{160\text{ meters per minute}}$ .

The vertical distance between Bob's head and the streetlight is 2.2 metres, and the vertical distance between Bob's head and the ground is 1.8 metres.

When Bob has moved for one minute, therefore 88 metres, his shadow will have moved an additional 88 x (1.8/2.2) = 72 metres using similar triangles.

88+72 = 160 metres, therefore 160m/min

Using similar triangle theory from this link

$\frac{4}{1.8} = (\frac{y}{y-x})$

1.1y = 2 x

differentiating the equation we get

1.1 dy = 2 dx

we know, dx = 88 m/min

hence $\boxed{dy = 160}$

We just conserve the angular velocity about the top point of the lampost::

x/4=88/2.2

get x = 160

Let the base of the streetlight be $A$ and the streetlight itself be $B$ . Let Bob's feet be $D$ and his head be $E$ . We are given $AB=4$ and $DE=1.8$ . Notice that line $AD$ represents the ground. The intersection of $BE$ and $AD$ is the shadow of Bob's head, which we will call $C$ . Triangles $ABC$ and $DEC$ are similar, so $CD/DA=\frac{9}{11} \implies CA=\frac{20}{11}DA$ .

Bob's speed is $\frac{d(DA)}{dt}=88$ . The speed of his head's shadow is: $\frac{d(CA)}{dt}=\frac{d(\frac{20}{11}DA)}{dt}=\frac{d( DA)}{dt}*\frac{20}{11}=88*\frac{20}{11}=\fbox{160}$