Question

A body starts from rest and is uniformly accelerated for 30 s. The distance travelled in the first 10 s is x1, next 10 s is x2 and the last 10 s is x3. Then x1:x2:x3 is the same as

A. 1:2:4
B. 1:2:5
C. 1:3:5
D. 1:3:9

Answer 1

The distance traveled in the first 10 seconds, next 10 seconds, and last 10 seconds when a body starts from rest and is uniformly accelerated can be determined using the equation d = ut + 0.5at^2. The ratio of the distances is 1:1:1.

Step-by-step explanation:

The distances traveled in the first 10 seconds, next 10 seconds, and last 10 seconds can be determined using the equation d = ut + 0.5at^2, where d is the distance, u is the initial velocity which is 0 since the body starts from rest, t is the time, and a is the acceleration.

For the first 10 seconds, the distance is x1 = 0(10) + 0.5a(10)^2 = 50a. For the next 10 seconds, the distance is x2 = 0(10) + 0.5a(10)^2 = 50a. And for the last 10 seconds, the distance is x3 = 0(10) + 0.5a(10)^2 = 50a.

This means that x1:x2:x3 = 50a : 50a : 50a. The ratio simplifies to 1:1:1, so the correct answer is A. 1:1:1.