1.find the distance travelled by a train in the first minute of its journey if it started from rest and has had constant acceleration of 0.5m/s. with explanation and step please

The distance of the traveled is 375 m.

Explanation:

Formula to be use for Motions Along a Straight Line

v=v_{0} +atv=v0+at          equation 1

x=v_{0} t + \frac{1}{2} at^2x=v0t+21at2     equation 2

where

vv     is the final velocity, unit is in m/s

v_{0}v0    is the initial velocity, unit is in m/s

xx     is the distance, unit is in meters (m)

aa     is the acceleration, unit is in m/s^2s2

tt      is the time duration of motion, unit is in seconds (s)

For the given information

v_{0}=0v0=0     because the car starts from rest

v=50m/sv=50m/s

t=15st=15s

Solving the problem

1. Using equation 1, solve for the acceleration

v=v_{0} +atv=v0+at

50 m/s = 0 + a (15 s)

Divide both sides of the equation by 15 s, we get:

a = \frac{50 m/s}{15s}15s50m/s

a = 3.33 m/s^2s2

2. Using equation 2, substitute the value of the acceleration then solve for x

x=v_{0} t + \frac{1}{2} at^2x=v0t+21at2

x =0+\frac{1}{2} (3.33m/s^2)(15s)^2x=0+21(3.33m/s2)(15s)2

x = 375 m

Therefore, the distance the car has traveled is 375 m within 15 seconds.

To learn more about this topic, just click on the following links:

Definition of acceleration

Additional examples

#LetsStudy

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)