A train moves with a uniform velocity of 36 $kmh^{-1} for 10s$. Find the distance traveled by it. (100 m)
Solution:
Velocity = V = $36 kmh^{-1}$ = $36 \times \frac{1000}{60 \times 60}$ =$\frac{36000}{3600}$ = $10 ms^{-1}$
Time t = 10 s
Distance = S =?
S = Vt
S = $10 \times 10$ = 100 m
A train starts from rest. It moves through 1 km in 100 s with uniform acceleration. What will be its speed at the end of 100 s.? $(20 ms^{-1})$
Solution:
Initial velocity Vi = 0 $ms^{-1}$
Distance S = 1 km = 1000 m
Time = 100 s
Final velocity Vf =?
S = $Vi \times t +\frac{1}{2} \times a \times t2$
1000 = $0 \times100 + \frac{1}{2} \times a \times (100)^{2}$
1000 = $\frac{1}{2} \times 10000a$
1000 = 5000a
a = 1000/5000 = $0.2 ms^{-2}$
\Now using 1st equation of motion
Vf = Vi + at
Vf = $0 + 0.2 \times 100$
Vf = $20 ms^{-1}$
A car has a velocity of $10 ms^{-1}$. It accelerates at $0.2 ms^{-2}$ for half a minute. Find the distance traveled during this time and the final velocity of the car. $(390 m, 16 ms^{-1})$
Solution: