A body starts from rest and accelerate for 4s until it reaches a velocity of 12m/s. It maintains the uniform velocity for another 5s. Calculate the acceleration and the total distance covered by the body
Answers
Answered by
1
Answer:
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
0
Answer:
a=3m/s and distance =84m
Explanation:
u=0
v=12m/s
t=4s
a=?
then,v=u+at
12=0+a(4)
a=3m/s
distance in this time
D1=a(t)^2/2
D1=3×8=24m
distance travelled in next 5s = vt=D2
12(5)=D2
60=D2
so, D=D1+D2
D=84m
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