Question

A body moves with uniform acceleration and travels 2 m in the first two seconds and 2.2 m in the next four seconds. What will be the distance travelled in the 8 th second of its motion?​

Answer 1

Answer:

A to B

2=u×2+  

2

1

​

×a×2×2⇒1=u+a

A to C

4.20=u×6+  

2

1

​

×a×6×6⇒0.7=u+3a

Let the Body travel from A to B in 2 s for a distance of 2 m

Let the Body travel from B to C for next 4s for a distance of 2.20 m

Velocity after 9 s=?

For AtoB

WKT, S=ut+  

2

1

​

at  

2

⇒2=2u+  

2

1

​

a∗2∗2⇒1=u+a...(1)

For B to C

WKT, S=ut+  

2

1

​

at  

2

⇒4.20=6u+  

2

1

​

a∗6∗6⇒0.7=u+3a...(2)

From (1) and (2), we get,

2a=−0.3⇒a=−0.15,−ve as it is decreasing acceleration .

u=1−a⇒u=1+0.15=1.15

Now, velocity at t=9s=v=u+at⇒v=1.15−0.15∗9=1.15−1.35=−0.2m/s

velocity is negative as it is decreasing

Explanation: