Given :
Initial speed , u = 0 m/s .
Final speed , v = 91 km/h = 25.28 m/s .
To Find :
a) Average acceleration .
b ) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.3 s .
Solution :
a )
We know ,by equation of motion :
b)
Also , by equation of motion :
Hence , this is the required solution .
Answer:
i dont really know what it is
v= s/t
Explanation:
250 km/ h =69.44m/s
S1=2 times 69.44 ≈ 139m
Next 2.5 seconds:
S2 = 100m
Average speed:
v=139m+100m/2s+2.5s = 239/4.5s = 53.2 m/s=192km/h
option E
given,
diameter = 4 mm
shutter speed = 1/1000 s
diameter of aperture = ?
shutter speed = 1/250 s
exposure time to the shutter time
N is the diameter of the aperture and t is the time of exposure
now,
inserting all the values
N₂² = 4
N₂ = 2 mm
hence , the correct answer is option E