Question

A car accelerates from rest at 1.0 m/s2 for 20.0 second along a straight road. It then moves at a constant speed for half an hour and then decelerates uniformly to a stop in 30.0 s. Find the total distance covered by the car and the average velocity of the car for the entire trip.

Answer 1

Total distance = 36500 m

The average velocity = 19.73 m/s

Further explanation

Given

vo=initial velocity=0(from rest)

a=acceleration= 1 m/s²

t₁ = 20 s

t₂ = 0.5 hr = 1800 s

t₃= 30 s

Required

Total distance

Solution

State 1 : acceleration

$\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m$

$\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s$

State 2 : constant speed

$\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m$

State 3 : deceleration

$\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)$

$\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m$

Total distance : state 1+ state 2+state 3

$\tt 200 + 36000 + 300=36500~m$

the average velocity = total distance : total time

$\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s$