An electric train moving at 20kgh-1 accelerate to a speed of 30kgh-1 in 20s.find the distance travel in meters during the period of acceleration.
kg is mass, not speed.
the acceleration is (10km/hr)/(20s) = 0.5 km/hr/s
s = vt + 1/2 at^2
= 20km/hr * 20s + 0.25 km/hr/s * (20s)^2
= 400 km-s/hr + 100km-s/hr
= 500 km-s/hr
= 500km * 1000m/km * 1s/3600s = 138.9 m
Or, you could convert everything to meters and seconds to start with, but that's too mundane.
Does Kgh-1 mean kg/hr. Do you mean km/h?
To find the distance traveled during the period of acceleration, you can use the equation:
distance = initial velocity * time + 0.5 * acceleration * time^2
In this case, the initial velocity (u) is 20 kgh^-1, the final velocity (v) is 30 kgh^-1, and the time (t) is 20 s. We need to find the acceleration (a) first.
Using the formula:
acceleration = (final velocity - initial velocity) / time
acceleration = (30 kgh^-1 - 20 kgh^-1) / 20 s
acceleration = 10 kgh^-1 / 20 s
acceleration = 0.5 kgh^-1 s^-1
Now plug the values into the distance formula:
distance = (20 kgh^-1 * 20 s) + 0.5 * (0.5 kgh^-1 s^-1) * (20 s)^2
distance = 400 kgh^-1s + 0.5 * 0.5 kgh^-1s^-1 * 400 s^2
distance = 400 kgh^-1s + 0.5 * 0.5 kgh^-1s^-1 * 160000 s
distance = 400 kgh^-1s + 0.25 kgh^-1s^-1 * 160000 s
distance = 400 kgh^-1s + 40000 kgh^-1s
distance = 40400 kgh^(-1)*s
Therefore, the distance traveled during the period of acceleration is 40,400 meters.