Question

The speed of a car was 50km/hr for the first 900sec then 40km/hr for the next 50km and then the car decelerated uniformly at 10km/h2 till it comes to rest. What is the average speed of the car?

Answer 1

Average speed is the fraction of total displacement travelled by the body to total time of travel.
V avg = total displacement/time.
# For first 900 sec, displacement=velocity×time
=50×900/(60×60)
=450/36 Km.
#For next 50 Km time travelled=displacement/velocity
=50/40
=5/4 hrs.
#For next time 40 Km/hr acts as initial velocity,where final velocity is given as 0.
using V^2 - U^2 = 2as
0 - 40^2 = 2 ×-10×s
s=40×2
s=80 Km.
using V-U =a×t
then time = 40/10
=4 hrs.

total displacement=(450/36)+50+ 80
=142.5 Km
total time= (900/60×60)+(5/4)+4
=5.5 hrs
average velocity=142.5/5.5
=26 Km/hrs.

Answer 2

Answer:

25.9 km/hr is the required average speed of the car .

Explanation:

Average speed = $\frac{Total \ distance\ travel }{time \ taken}$

Let $v_{1} , v_{2}, v_{3}$ be the velocity , $x_{1} , x_{2} , x_{3}$ be the distances and

let $t_{1} , t_{2} , t_{3}$ be the times .

Therefor   from the question we have ,

In 900sec the speed of car was 50 km/hr

$v_{1}$ = 50 km/hr

$t_{1}$ = 900 sec ⇒ $t_{1} = \frac{900}{ 60 . 60 } = \frac{1}{4} hrs$  = 0.25 hr

∴$x_{1} = v_{1} . t_{1}$ ⇒  $x_{1}$  = 50 ×$\frac{1}{4}$ km = 12.5km

For the next 50km speed of car is  40km/hr

$v_{2}$ = 40 km/hr                  

$x_{2}$ = 50 km    

∴$t_{2}$ = $\frac{x_{2} }{v_2}$ = $\frac{50}{40}$ = $\frac{5}{4}$ hr  = 1.25 hr

For the next  40 km /hr act as initial velocity , where final velocity is  zero

Therefore from the formula v = u + at

⇒ v = u + $a_{3} t_{3}$

⇒ v - u =  $a_{3} t_{3}$

⇒  (0 - 40) km/hrs =  - 10$hr^{2}$× $t_{3}$

⇒$t_{3}$ = 4 hr

Now , from the equation of motion  v2 = u2 + 2as

Where v is the final velocity , u is the initial velocity , a is acceleration and s is distance.

⇒ v2 = u2 + 2as

⇒   v2 - u2 =  2as     ⇒ 0 - $(40 )^{2}$ = 2 × -10 ×  s

⇒s =  $\frac{1600}{20} = 80$ km  = $x_{3}$

∴ Average speed =     $\frac{x_{1} + x_{2} + x_{3}}{t_{1} + t_{2} + t_{3}}$ = $\frac{12.5 +50 +80}{0.25 +1.25 + 4 }$  = $\frac{142.5}{5.5} = 25.9$ km/hrs

Hence , the average speed of the car is 25.9 km/hr

#SPJ2