PHYSICS HIGH SCHOOL

A train starting from a distance of velocity of 90 km.h-1 in 10 minutes assuming that the acceleration is uniform find the acceleration and the distance travelled by the train for attending the velocity

Answers

Answer 1

Answer:

To make this problem solvable and you can get the help you need, I'll complete and arrange some data.

Answer:

Acceleration: $0.0417\ m/s^2$ , Distance=7,500 m

Explanation:

Uniform Acceleration Motion

It's a type of motion in which the velocity of an object changes uniformly over time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

$v_f=v_o+at\qquad\qquad [1]$

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+(a.t^2)/(2)\qquad\qquad [2]$

Using the equation [1] we can solve for a:

$\displaystyle a=(v_f-v_o)/(t)\qquad\qquad [3]$

The problem will be rewritten as follows:

A train starting from rest reaches a velocity of 90 km/h in 10 minutes. Assuming that the acceleration is uniform, find the acceleration and the distance traveled by the train for attending the velocity.

Let's take the relevant data:

vo=0

vf=90 Km/h*1000/3600 = 25 m/s

t = 10 minutes = 10*60 = 600 seconds

Now compute the acceleration by using [3]:

$\displaystyle a=(25-0)/(600)=0.0417$

$a=0.0417\ m/s^2$

Finally, compute the distance:

$\displaystyle x=0*600+(0.0417\cdot 600^2)/(2)$

$x=7,500\ m$

Note: We used the value of the acceleration with more precision than shown.

Acceleration: $\mathbf{0.0417\ m/s^2}$ , Distance=7,500 m

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