A body accelerates uniformly from 10 m/s to 25
m/s in 5 s. Find [3]
(i) The velocity of the body after next five second.
(ii) The distance covered by the body in that time.
Answers
Answer:
625
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
Given:
Initial velocity, u = 10 m/s
Final velocity, v = 25 m/s
Time = 5 s
To Find:
(i) The velocity of the body after next five second.
(ii) The distance covered by the body in that time.
Calculation:
- Applying 1st equation of motion:
v = u + at
⇒ 25 = 10 + 5a
⇒ a = 3 m/s²
(i) After next 5 second, T = 5 + 5 = 10 s
- Again using 1st equation of motion:
V = u + at
⇒ V = 10 + (3 × 10)
⇒ V = 40 m/s
(ii) For the next 5 seconds, U = 25 m/s
Using 3rd equation of motion, we get:
V² = U² + 2aS
⇒ S = (V² - U²) / 2a
⇒ S = (40² - 25²) / (2 × 3)
⇒ S = (1600 - 625) / 6
⇒ S = 975/6
⇒ S = 162.5 m
- So,
(i) The velocity of the body after next five seconds is 40 m/s.
(ii) The distance covered by the body in that time is 162.5 m.