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a=4 m/s²
t=10s
u=0m/s
v=u+at. -1st eq. of motion
v=0+40m/s
v=40m/s
v²=u²+2as -3rd eq of motion
(40)²=2×4s
s= 40×40/8
s=40×5=200m
t=10s
u=0m/s
v=u+at. -1st eq. of motion
v=0+40m/s
v=40m/s
v²=u²+2as -3rd eq of motion
(40)²=2×4s
s= 40×40/8
s=40×5=200m
Answered by
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At the start,
initial velocity, u = 0
acceleration, a = 4m/s²
time taken, t = 10s
Distance covered, s = ut + ¹/₂ at²
= u(0) + ¹/₂ (4) (10)²
= ¹/₂ (400)
= 200 metres
Thus, the distance covered in 10 seconds is 200 metres
Hope it helps!
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