Question

a truck is moving with the constant acceleration of 25km/h2 in a fixed direction speeds up from 10km/h to 15km/h . find the time taken by the truck to change its speed . also find the distance covered.

Answer 1

Hii

v = u + at

15 = 10 + 25t

25t = 5

t = 5÷25

t = 1/5 hr

DISTANCE :

v²-u²=2as

(15)²-(10)²=2×25×s

225-100 = 50s

s = 125/50

s = 2.5 km

Answer 2

Here,truck's uniform acceleration,a=25km/h^2

=(25000/3600)m/s^2

=(250/36)m/s^2

inertial velocity,u=10km/h

=(10000/3600)m/s

=(25/9)m/s

final velocity,v=15km/h

=(15000/3600)m/s

=(25/6)m/s

As the first equation of motion,

v=u+at

or,25/6=25/9+(250/36)t

or,(25/6)-(25/9)=(250/36)t

or,25/18=(250/36)t

or,(25/18)/(250/36)=t

or,(25/18)*(36/250)=t

or,1/5=t

so,t=0.2s

That means,it will change it's velocity from 10-15 km/h within 0.2 seconds.

Also,

v^2=u^2+2as

or,(25/6)^2=(25/9)^2+2*(250/36)s

or,625/36=(625/81)+(500/36)s

or,(625/36)-(625/81)=(500/36)s

or,3125/324=(500/36)s

or,(3125/324)/(500/36)=s

or,(3125/324)*(36/500)=s

so,s=0.694m(approximately)

Ans.t=0.2s,s=0.694m.