Solveeit Logo

Question

Physics Question on Motion in a straight line

A body covers 26, 28, 30, 32 meters in 10th10^{th}, 11th11^{th}, 12th12^{th} and 13th13^{th} seconds respectively. The body starts

A

from rest and moves with uniform velocity

B

from rest and moves with uniform acceleration

C

with an initial velocity and moves with uniform acceleration

D

with an initial velocity and moves with uniform velocity

Answer

with an initial velocity and moves with uniform acceleration

Explanation

Solution

The distance covered in nth second is Sn=u+12(2n1)aS_n = u + \frac{1}{2} (2n - 1)a where u is initial velocity & a is acceleration then 26=u+19a226 = u + \frac{19a}{2} .....(1) 28=u+21a228 = u + \frac{21a}{2} ....(2) 30=u+23a230 = u + \frac{23 a}{2} .....(3) 32=u+25a232 = u + \frac{25a}{2} ......(4) From e (1) & (2) we get u = 7m/sec,a=2m/sec27 m/ \sec, a = 2 \, m/ \sec^2 \therefore The body starts with initial velocity u = 7 m/sec and moves with uniform acceleration a=2m/sec2a = 2 m/ \sec^2