. A particle starts from rest and travels with a constant acceleration of 4m per sec² for some time. After that it starts de accelerating at 3 m/s² and finally comes to rest. If during the motion, it covers a total distance of 42 meters, then what is the total time of motion?
I need the explanation with steps please
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Answer:
Explanation:
Displacement in first, 10sec
s
1
=
2
1
a
1
t
1
2
=
2
1
×2×10
2
=100m
Speed achieve in first 10sec
v
1
=u
1
+a
1
t
1
=0+2×10=20m/s
Displacement in 30sec, acceleration is zero.
s
2
=v
1
t
2
=20×30=600m
Displacement in last due to deacceleration, a
3
=−4m/s
2
v
2
−v
1
2
=2as
s
3
=
2×(−4)
0−20
2
=50m
Net displacement
S
total
=s
1
+s
2
+s
3
=100+600+50=750m
Total displacement is 750m
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