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A 1000 kg vehicle moving with a speed of 20m/s is brought to rest in a distance of 50 meters :
(i) Find the acceleration.
(ii)Calculate the unbalanced force acting on the vehicle.
Given,
Initial velocity (u) = 20m/s
Final velocity (v) = 0m/s
Distance moved (s) = 50m
We know,
${\mathrm{v}}^{2}={\mathrm{u}}^{2}+2\mathrm{as}\phantom{\rule{0ex}{0ex}}$
$0={20}^{2}+2\mathrm{a}\times 50\phantom{\rule{0ex}{0ex}}$
$0=400+100\mathrm{a}\phantom{\rule{0ex}{0ex}}$
$-400=100\mathrm{a}\phantom{\rule{0ex}{0ex}}$
$\mathrm{a}=-\frac{400}{100}\phantom{\rule{0ex}{0ex}}$
$\mathrm{a}=-4\mathrm{m}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}$
Now,
$\mathrm{Force(F)=mass(m)\times acceleration(a)}\phantom{\rule{0ex}{0ex}}$
$\mathrm{F}=1000\times (-4)\phantom{\rule{0ex}{0ex}}$
$\mathrm{F}=-4000\mathrm{N}$
Hence, the unbalanced force acting on the vehicle is -4000N.
Updated on: 10-Oct-2022
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