Question

An unloaded car moving with velocity $u$ on a frictionless road can be stopped in a distance $s$. If passengers add $40$% to its weight and breaking force remains the same, the stopping distance at velocities is now

• A. $1.4s$
• B. $\sqrt{1.4}s$
• C. ${\left(1.4\right)}^{2}s$
• D. $\frac{1}{1.4}s$

Answer 1

The correct option is A $1.4s$

Let $m$ be the mass of the unloaded car. Then,

$\frac{1}{2}m{u}^2=FS$----------(i)

where $F=$ retarding force.

When $40\%$ weight is added, new mass is given by:

$m_1=m+\left(\frac{40}{100}\right)m=1.4m$

Now, $\frac{1}{2}{m}_1{u}^2=F{S}_1$

or, $\frac{1}{2}\times 1.4m{u}^2=F{S}_1$-------(ii)

So, from (i) and (ii), we can write:

$S_1=1.4S$