Question

- A body starts from rest and moves with uniform acceleration of 0.25 m/s2. Find the distance
travelled by it in 10 s and its velocity. Find also the distance covered by it in the 10th second of
its motion.
(12.5 m; 2.5 m/s; 2.375 m)

Answer 1

Explanation:

u = 0m/s

a = 0.25m/s

t = 10s

S = ut + 1/2at²

S = 12.5m

v = u + at

v = 0.25 * 10

v = 2.5m/s

Snth = u + 1/2a(2n - 1)

S10th = 1/2*0.25*19

S10th = 2.375m

Answer 2

Explanation:

u=0ms

a=0.25m/s2

t=10sec.s

formula required to find distance:

s=ut + 1/2at^2

0×10+1/2 × 0.25 ×10×10

0+12.5

12.5 =distance

formula required to find velocity:

2as=v^2-u^2

2×0.25×12.5=v^2-0

6.25=v^2

v=2.5

so, there the given time is 1 sec

formula required : s =ut +1/2at^2

s=0×1+1/2×0.25×1×1

s=0+0.125

s=0.125