An object is having a velcoity 4.0m/s is accelerated at the rate of 1.2m/s2 for 5 secs. Find the distance travelled during the period of acceleration. Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics
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4
Solution :
Initial velocity=u=4 m/s
acceleration=a=1.2m/s2
time=t=5sec
Distance =s=?
Formula to be used : s=ut +1/2 at²
s=4x5 +1/2 x 1.2 x 5²
=20 +12x25/20
=20+ 15
=35m
∴Distance travelled is 35m
Initial velocity=u=4 m/s
acceleration=a=1.2m/s2
time=t=5sec
Distance =s=?
Formula to be used : s=ut +1/2 at²
s=4x5 +1/2 x 1.2 x 5²
=20 +12x25/20
=20+ 15
=35m
∴Distance travelled is 35m
Answered by
1
heya.......
Initial velocity u=4.0 m/s² acceleration a=1.2 m/s².
time t=5.0 s
from the equation of accelerated motion on a straight line ,
x=ut+½at²
x=4x5+½x1.2x5²
x =20+15= 35 m
tysm...#gozmit
Initial velocity u=4.0 m/s² acceleration a=1.2 m/s².
time t=5.0 s
from the equation of accelerated motion on a straight line ,
x=ut+½at²
x=4x5+½x1.2x5²
x =20+15= 35 m
tysm...#gozmit
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