A body starts from rest and moves with constant acceleration 10m/sec². How much distance will it travel in the 4th sec of it's motion.
Answers
Answered by
1
Hey good evening siso I m ur foll ow er. ur ans
The total distance travelled is
=
625
m
Explanation:
The initial velocity is
u
=
0
m
s
−
1
The acceleration is
a
=
10
m
s
−
2
The time is
t
=
5
s
Apply the equation of motion
v
=
u
+
a
t
The velocity is
v
=
0
+
10
⋅
5
=
50
m
s
−
1
The distance travelled in the first
5
s
is
s
=
u
t
+
1
2
a
t
2
=
0
⋅
5
+
1
2
⋅
10
⋅
5
2
=
125
m
The distance travelled in the following
10
s
is
s
1
=
v
t
1
=
50
⋅
10
=
500
m
The total distance travelled is
d
=
s
+
s
1
=
125
+
500
=
625
m
Answered by
6
Answer:
35m
Explanation:
u = 0
a = 10m/s²
For 3 seconds
s = 0(3) + ½(10)(3)²
= 5(9)
= 45 m
For 4 seconds
s = 0(4) + ½(10)(4)²
= 5(16)
= 80m
Distance travelled in 4th second = 80m - 45m
= 35m
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