Question

# a bus starting from rest attains a speed of 36 km per hour in 5 seconds calculate acceleration and distance travel by bus during this duration

Open in App
Solution

## Dear Student, $\mathrm{initial}\mathrm{velocity},\mathrm{u}=0\phantom{\rule{0ex}{0ex}}\mathrm{final}\mathrm{velocity},\mathrm{v}=36\mathrm{km}/\mathrm{h}\phantom{\rule{0ex}{0ex}}\mathrm{v}=36×\frac{5}{18}\mathrm{m}/\mathrm{s}=10\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{time},\mathrm{t}=5\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{v}=\mathrm{u}+\mathrm{at}\phantom{\rule{0ex}{0ex}}10=0+\mathrm{a}×5\phantom{\rule{0ex}{0ex}}\mathrm{a}=2\mathrm{m}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{v}}^{2}-{\mathrm{u}}^{2}=2\mathrm{as}\phantom{\rule{0ex}{0ex}}{10}^{2}-{0}^{2}=2×2\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{s}=\frac{100}{4}=25\mathrm{m}$

Suggest Corrections
0
Related Videos
Third Equation of Motion
PHYSICS
Watch in App
Explore more