Question

the train trawler at a speed of 25m/a ,brake are applied to stop the train at the acceleration of 0.5 m per square second. the distance traveled by the train in
a) 500m
b) 750
c) 625m
d) none

please fast ​

Answer 1

$\boxed{\bold{\large{GIVEN:}}}$

• Initial velocity (u) = 25ms⁻¹.
• Final velocity (v) = 0ms⁻¹ [Because breaks are applied].
• Acceleration (a) = -0.5ms⁻² [It is negative because it is acting in the opposite direction]

$\boxed{\bold{\large{TO \:\: FIND:}}}$

• The distance (s) of the train.

$\boxed{\bold{\large{SOLUTION:}}}$

Recall the three equations of motion:

$\boxed{\substack{\displaystyle v=u+at \\ \displaystyle s=ut+\frac{1}{2}at^2 \\ \displaystyle 2as=v^2-u^2}}$

It is apt to use the third formula because v, u and a is given and we need to find s.

$\sf{\red{2as=v^2-u^2}}$

Substitute.

$\sf 2 \times (-0.5) \times s=0^2-25^2$

$\sf \not{-}s=\not{-}625$

$\boxed{\sf{\blue{s=625 m.}}}$

∴DISTANCE IS 625 m.