Question

A train starts from rest. It moves through 1km in 100s with uniform acceleration. What will be its speed at the end of 100 s?

Answer 1

Explanation:

Using the equations of motion,

s = ut+1/2 at^2

v = u + at,

where

s = displacement,

u = initial velocity

a = acceleration

t = time

Using the first equation s = ut+1/2 at^2

1ooom = 0(t) +1/2(a)(100s)^2

Solving for a, we obtain:

a = 0.2m/s^2

Using the other equation of motion:

v = u + at,

v = 0 + (0.2m/s^2)(100s)

v = 20m/s

Answer 2

Answer:

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Explanation:

Initial velocity(vi)

= 0 m/s

s = 1 km

convert into m

s = 1000 m

t = 100 sec

to find

Acceleration = ?

Final velocity(vf) = ?

using formula

s = vit + 1/2 x at^2

1000 = (0) (100) + 1/2 x a(100)^2

1000 = 0 + 1/2 x a(100)^2

1000 = 0 + 1/2 x a(10000)

1000 = a × 5000

a = 5000/1000

a = 0.2ms^-1

Now,

vf = vi + at

vf = 0 + (0.2) × 100  

vf = 20ms^-1