Question

An object start from rest and is uniformly accelerated of 30m/s . Find the ratio of distance traveled by object in of time interval of 10s each

Answer 1

The distance covered in first 10 sec is x1 , for next 10 sec is x2and for last 10 sec is x3 the then the ratio of X1:X2:X3
Answer:

k=0,1,2,3,....
1:3:5:7:(2k+1)

Explanation:

please look at the animation and calculate the colored areas.


k=0,1,1,3,4...
1:3:5:7:(2k+1)



A08

Apr 12, 2017

Answer:

x1:x2:x3=1:3:5

Explanation:

The applicable kinematic expression is 
s=ut+12at2
s is distance traveled in time t, u is initial velocity and a is acceleration.

Distance traveled in first 10s
x1=0×10+12a×102
⇒x1=50a

Distance traveled in next 10s
x2=s20−s10=(0×20+12a×202)−50a
⇒x2=200a−50a=150a

Distance traveled in last 10s
x3=s30−s20=(0×30+12a×302)−200a
⇒x3=450a−200a=250a

the required ratio is 
x1:x2:x3=50a:150a:250a
⇒x1:x2:x3=1:3:5