Calculate Th e acceleration and distance of the body moving with 5m/s which comes to rest after travelling for 6 seconds
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acceleration = final velocity-initial velocity ÷time taken...
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using Newton's first law of kinematics.
a= (v-u)/t
a= (5-0)/
a= 5/6 m/s^2
for distance we use Newton's second law of kinematics.
s= ut + 1/2×at^2
s= 5×6 + 1/2×(-5/6)6×6
{ here -ve sign indicates that body is regarding}
s= 30 - 30/2
s= 30 - 15
s= 15m.
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a= (v-u)/t
a= (5-0)/
a= 5/6 m/s^2
for distance we use Newton's second law of kinematics.
s= ut + 1/2×at^2
s= 5×6 + 1/2×(-5/6)6×6
{ here -ve sign indicates that body is regarding}
s= 30 - 30/2
s= 30 - 15
s= 15m.
hope u liked it
please mark it as brainliest. :)
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