A body travels 20cm in first 0.2 sec and 22cm in next 0.4 sec .what will be velocity at the end of 0.7th sec from start
Answers
velocity at the end of 0.7sec from start would be 10cm/s
let initial velocity is u and acceleration is a.
case 1 : s₁ = 20 cm and t₁ = 0.2 sec
using formula, s₁ = ut₁ + 1/2 at₁²
⇒(20cm) = u × 0.2 + 1/2 × a × (0.2)²
⇒0.2 = 0.2u + 0.2 × 0.1 × a
⇒1 = u + 0.1a
⇒10 = 10u + a ............(1)
and velocity after 0.2 sec, v = u + at₁
= u + 0.2a
case 2 : s₂ = 22cm and t₂ = 0.4 sec
using formula, s₂ = vt₂ + 1/2 at₂²
⇒(22cm)² = (u + 0.2a) × (0.4) + 1/2 × a × (0.4)²
⇒0.22 = 0.4u + 0.08a + 0.08a
⇒0.22 = 0.4u + 0.16a
⇒11 = 20u + 8a ...........(2)
from equations (1) and (2) we get,
11 - 20 = 8a - 2a
⇒a = -9/6 = -3/2 = -1.5 m/s²
u = (10 - a)/10 = 11.5/10 = 1.15 m/s
velocity at the end of 0.7sec, v = u + a(0.7sec)
= 1.15 - 1.5 × 0.7
= 1.15 - 1.05
= 0.10
= 10cm/s
Answer:
10cm/s
Explanation:
case(I)
Sn=20cm= 0.20m , Tn=0.2 sec
Sn=ut+1at²
2
0.2 = 0.2u+a (0.2)²=0.2u+a 0.04
2. 2
0.2 = 0.2u+0.02a
divide eqn by 2
0.1 = 0.1u+0.01a ——— eqn 1
case (ii)
Sn=0.20m+0.22m=0.42m , Tn=0.2+0.4=0.6 s
0.42 = 0.6u+0.18a
divide eqn by 6
0.06= 0.1u+ 0.01a ———eqn 2
from eqn 1 and 2
we get,
a= -1.5 m/s²
u= 1.15 m/s
velocity at 7th sec
v = u + at
v = 1.15 + -1.5*0.7
v = 10 cm/s