# How do I find the speed of a platform with respect to the ground when a person walks over it?

#### Chemist116

The problem is as follows:

A man whose mass is $60\,kg$ is standing in the rear end of a platform whose mass is $140\,kg$ which is moving with no friction over an ice rink with a velocity of $4\,\hat{i}\,\frac{m}{s}$. The man moves with a speed of $2\,\frac{m}{s}$ with respect to the platform. Find the speed of the platform with respect to the ground in $\frac{m}{s}$.

The alternatives given are as follows:

$\begin{array}{ll} 1.&-4.6\,\hat{i}\,\frac{m}{s}\\ 2.&-3.4\,\hat{i}\,\frac{m}{s}\\ 3.&2\,\hat{i}\,\frac{m}{s}\\ 4.&3.4\,\hat{i}\,\frac{m}{s}\\ 5.&4.6\,\hat{i}\,\frac{m}{s}\\ \end{array}$

The only thing which I could find was the speed of the man:

v_man-v_platform = 2

v_man=2+v_platform = 2+ 4 = 6

But I'm not sure how to relate the other given information. Can someone help me here?.

#### skeeter

Math Team
$p_0 = p_f$

$(140+60) \cdot 4i = 140v_f + 60(v_f+2i)$

#### Chemist116

$p_0 = p_f$

$(140+60) \cdot 4i = 140v_f + 60(v_f+2i)$
Thanks for that. Would you mind helping me with the problem about the person walking in a boat? I'm stuck exactly at how should I use the information of the length.

#### Chemist116

Thanks for that. Would you mind helping me with the problem about the person walking in a boat? I'm stuck exactly at how should I use the information of the length.
@skeeter I'm revisiting this problem how did you arrived to $v_{\textrm{man}}=v_f+2$

When it mentions that the man moves with respect of the platform 2i would it be like this:

$v_{f}-v_{\textrm{man}}=2$

Hence:

$v_{\textrm{man}}=v_{f}-2$

This is the part where I'm stuck. Can you help me clearing out this idea?

#### skeeter

Math Team
the man walks 2 m/s relative to the platform

the platform moves at a velocity $v_f$ to the ice rink

therefore, the man moves at a velocity $(v_f+2)$ m/s relative to the ice rink

topsquark