a body start with the initial velocity of 10ms-1 and acceleration 5ms-2 find the distance covered by the 5 sec
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using s=ut+1/2at²
s=10×5+1/2×5×5²
s=50+1/2×5×25
s=50+1/2×125
s=50+62.5
s=112.5m
distance = 112.5m
Hope it helps:)
Brainliest answer pls
s=10×5+1/2×5×5²
s=50+1/2×5×25
s=50+1/2×125
s=50+62.5
s=112.5m
distance = 112.5m
Hope it helps:)
Brainliest answer pls
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