Question

a train starts its journey from station P; accelerates at the rate of 2m/s2 and reaches its maximum speed in 10 seconds. it maintain its speed for 30 minutes and retards uniformly to rest at the station Q after the next 20 seconds. calculate a.the maximum speed of the train. b.retardation c.distance between P and Q.

Answer 1

Please adjust with the quality if pictures. Note:the minus sign in front if acceleration in 3rd case indicates retardation (negative acceleration)

Answer 2

Answer:

The maximum speed, retardation and the distance between P and Q are 20 m/s, 1 m/s² and 3.9 km.

Explanation:

Given that,

Acceleration $a =2 m/s^2$

Time t = 10 sec

(I). We need to calculate the maximum speed

Using equation of motion

$v = u+at$

Where, v = speed

u = initial speed

a = acceleration

t = time

Put the value in the equation

$v =0+2\times10$

$v = 20\ m/s$

(II). Given that,

Time t = 20 sec

We need to calculate the retardation

Formula of retardation is

$a = \dfrac{v-u}{t}$

Put the value into the formula

$a =\dfrac{0-20}{20}$

$a =-1\ m/s^2$

Negative sigh shows the retardation

(III). We need to calculate the distance between P and Q

We need to draw a figure,

From figure,

Distance = Area under the curve

$D=area PRT+area RSTU+areaUSQ$

$D = \dfrac{1}{2}\times20\times10+20\times180+\dfrac{1}{2}\times20\times20$

$D = 3900\ m$

$D = 3.9\ km$

Hence, The maximum speed, retardation and the distance between P and Q are 20 m/s, 1 m/s² and 3.9 km.