5. A particle covers 10m in first 5s and and 10m in next 3s. Assuming constant acceleration, find initialspeed, acceleration and distance covered in next 2s.
Answers
Case I
s=ut+
2
1
at
2
10=5u+
2
1
a(5)
2
20=10u+25a
4=2u+5a..............(1)
Case 2
In next 3s the particle covers more 10m distance. So
20=8u+
2
1
a(8)
2
5=2u+8a.........(2)
On solving equation (1) and (2)
4=2u+5a
5=2u+8a
a=
3
1
m/s
2
Put the value of a in equation (1)
u=
6
7
m/s
Now to find distance in next 10 s. total time will be 10s
s=
6
7
×10+
2
1
×
3
1
×(10)
2
s=28.33m
Distance travelled in next 2 sec
s=28.33−20=8.33m
Given :
Distance covered in first 5s = 10m
Distance covered in next 3s = 10m
To Find :
Initial velocity, acceleration and distance covered by particle in next 2s.
Solution :
Let initial velocity of particle be u and acceleration be a.
★ Particle covers 10m in first 5 seconds :
➠ 10 = 5u + 1/2 a × (5)²
➠ 10 = 5u + 25/2 a
➠ 20 = 10u + 25a
➠ 5(4) = 5(2u + 5a)
➠ 4 = 2u + 5a
➠ 2u = 4 - 5a .......... (I)
★ Particle again covers 10m in next 3 seconds :
Indirectly we can say that particle covers 20m in first 8 seconds.
➠ 20 = 8u + 1/2 a × (8)²
➠ 20 = 8u + 32a
➠ 4(5) = 4(2u + 8a)
➠ 5 = 2u + 8a
➠ 2u = 5 - 8a .......... (II)
On solving both equations, we get
⭆ 4 - 5a = 5 - 8a
⭆ -5a + 8a = 5 - 4
⭆ 3a = 1
⭆ a = 0.33 m/s² [A]
By substituting value of a in the first equation;
⭆ 2u = 4 - 5a
⭆ 2u = 4 - 5(0.33)
⭆ 2u = 4 - 1.65
⭆ 2u = 2.35
⭆ u = 1.175 m/s [B]
Now let's calculate last part of the question :)
In this part, we have to find distance covered by particle in last 2 seconds
Instead of solving by complex way, we can easily follow these steps :
- Calculate total distance travelled by particle 10 seconds.
- Subtract 20m from the result in order to get distance covered in last 2 second.
Distance covered in first 10 seconds :
⇒ S₁₀ = ut + 1/2 at²
⇒ S₁₀ = 1.175(10) + (0.33) (10)²/2
⇒ S₁₀ = 11.75 + 33/2
⇒ S₁₀ = 11.75 + 16.5
⇒ S₁₀ = 28.25 m
Distance covered in last 2 seconds :
➙ S = S₁₀ - S₈
➙ S = 28.25 - 20
➙ S = 8.25 m [C]