(a) state a condition under which a body such a way that the magnitude of its average velocity is equal to its average speed. (b) A train starting from rest moves with a uniform acceleration of 0.2m\ssquare of 5 minutes. calculate the final velocity and distance travelled in this time.
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a) when a body travels in a straight line it's displacement = distance
as displacement and distance are same speed and velocity are also same.
b) the starts from rest so initial velocity is equal to zero (u=0)
acceleration = 0.2 m/s^2
time = 5min = 300 seconds
from first equation of motion
v = u + at
so, v = 0 + 0.2 * 300
v = 60 m/s ........1
distance traveled is equal to
from second equation
s = ut + 1/2 at ^2
s = 0*300 + 1/2 * 0.2* 300 ^ 2
s = 1/2 *0.2 * 300 * 300 = 9000 m
therefore the answer is
v = 60m/s
s = 9000 m or 9 km
as displacement and distance are same speed and velocity are also same.
b) the starts from rest so initial velocity is equal to zero (u=0)
acceleration = 0.2 m/s^2
time = 5min = 300 seconds
from first equation of motion
v = u + at
so, v = 0 + 0.2 * 300
v = 60 m/s ........1
distance traveled is equal to
from second equation
s = ut + 1/2 at ^2
s = 0*300 + 1/2 * 0.2* 300 ^ 2
s = 1/2 *0.2 * 300 * 300 = 9000 m
therefore the answer is
v = 60m/s
s = 9000 m or 9 km
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