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3. Here, acceleration a = 0.1 ms-2
Time t = 2min = 2 x 60 = 120s
Initial speed u = 0
(a) From Ist equation of motion, speed acquired, v = u + at
= 0 + 0.1 x 120
= 12 ms-2
(b) From IInd equation of motion,
Distance travelled,
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= 0.1 x 60 x 120 = 720 m
5. Here, initial velocity, u = 0
Acceleration, a = 2 cm s-2
Time, t = 3 s
From, Ist equation of motion, v = u + at = 0 + 2 x 3 = 6 ms-1
1. Uniform acceleration a = 4 m/s²
Initial velocity u= 0 m/s
Time t = 10 sec.
Velocity v after 10 sec= u + a t = 0 m/s + 4 m/s² × 10 s =40 m/s.
We shall calculate the distance covered by all possible methods
Using Average Velocity:
Initial velocity u= 0 m/s
Final velocity v= 40 m/s
Average velocity= ½( u +v) = ½ ( 0+40)= 20 m/s
Time= 10 s
Distance covered in 10 s= av. vel. × time= 20m/s × 10s = 200 m
Using the relation s = u t+ ½ a t²
u = 0 m/s, a = 4 m/s², t= 10 sec.
s = 0×10 + ½ 4× 10² = 200 m
Using the relation v² - u² = 2 a s,
u = 0 m/s, v= 40 m/s, a= 4 m/s²
s = 1/2 a ( v² - u²)= 1/2×4 ( 40² - 0²) = 1/8(1600–0)= 1600/8= 200 m.
So all relations give the same results as it should.
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