Question

A train running at 30 m/s in slowed uniformly to a stop in 44 second find the acceleration and the stopping distance.

Answer 1

$\Huge\bold\red{Question}$

A train running at 30 m/s in slowed uniformly to a stop in 44 second find the acceleration and the stopping distance.

$\large\bold\green{Answer}$

u = 30 m/s,v= 0,t= 44s

a = v-u/t

= 0-30/44

a=-0.68 m/s^2

s= ut+1/2at^2

= (30)(44)+1/2(-0.68)(44)^2

=662m

@AngelicSparks

Answer 2

Given :-

Initial velocity of the train = 30 m/s

Final velocity of the train = 0 m/s

Time taken for the train to stop = 44 sec

To Find :-

The acceleration of the train.

The stopping distance of the train.

Analysis :-

Here we are given with the time taken, initial and final velocity of the train.

Firstly find the acceleration using the first equation of motion by substituting the given values in the question.

Then using the second equation of motion find the stopping distance accordingly.

Solution :-

We know that,

• u = Initial velocity
• t = Time
• v = Final velocity
• a = Acceleration

Using the formula,

$\underline{\boxed{\sf First \ equation \ of \ motion=v=u+at}}$

Given that,

Final velocity (v) = 0 m/s

Initial velocity (u) = 30 m/s

Time (t) = 44 sec

Substituting their values,

$\sf 0=30+a \times 44$

$\sf a=\dfrac{(0-30)}{44}$

$\sf a=\dfrac{-30}{44}$

$\sf a=-0.68 \ m/s$

Therefore, the acceleration of the train is -0.68 m/s.

We know that,

• u = Initial velocity
• t = Time
• s = Displacement
• a = Acceleration

Using the formula,

$\underline{\boxed{\sf Second \ equation \ of \ motion=s=ut+\dfrac{1}{2} at^2}}$

Given that,

Initial velocity (u) = 30 m/s

Acceleration (a) = -0.68 m/s

Time (t) = 44 sec

Substituting their values,

$\sf s=30 \times 44+\dfrac{1}{2} \times 0.68 \times 44^2$

$\sf s=1320+\dfrac{-1316.48 }{2}$

$\sf s=1320+-658.24$

$\sf s=661.76 \ m$

Therefore, the stopping distance is 661.76 m.