inframundosa921

2023-03-25

A truck on a straight road starts from rest, accelerating at $2m{s}^{-2}$ , until it reaches a speed of $20m{s}^{-1}$ . Then the truck travels at a constant speed for $20s$ until the brakes are applied causing the truck to decelerate uniformly to stop it in additional $5s$ .
(a) For how much time the truck is in motion?
(b) What is the average speed of the truck for its entire motion?

meakQueueksiu7e

Step 1: Given Data
A truck is accelerating while traveling on a straight road $a=2m{s}^{-2}$
the truck's initial velocity, $u=0m{s}^{-1}$
The final velocity of the truck, $v=20m{s}^{-1}$
The truck travels at a constant speed for a time, $t2=20s$
The time required to stop after brakes are applied, $t3=5s$
Step 2: Determine the total time for which the truck is in motion
Let the time it takes to attain $20m{s}^{-1}$ as the final velocity be $t1$ .
As we know that the first equation of motion is $v=u+at$
$⇒t1=\frac{v-u}{a}$
$⇒t1=\frac{20-0}{2}$
$⇒t1=10s$
Thus, the total time for which the truck is in motion is $t=t1+t2+t3=10+20+5=35s$
Step 3: Determine the distance covered by the truck for the first $10s$
As we know that the second equation of motion is $S=ut+\frac{1}{2}a{t}^{2}$
Let the distance traveled for the first $10s$ be $S1$
$⇒S1=0+\frac{1}{2}×2×{\left(10\right)}^{2}$
$⇒S1=100m$
Step 4: Determine the distance covered by the truck for the next $20s$
For next $20s$ the truck moved at a constant speed. Hence, $a=0m{s}^{-1}$
Let the distance traveled for the next $20s$ be $S2$
On applying the given data in $S=ut+\frac{1}{2}a{t}^{2}$ we get,
$⇒S2=20×20+0$
$⇒S2=400m$
Step 5: Determine the distance covered by the truck for the last $5s$
For the last $5s$ the truck gets stopped after applying brakes
Let the distance traveled for the last $5s$ be $S3$
On applying the equation, $S=\frac{1}{2}\left(v+u\right)t$
$⇒S3=\frac{1}{2}\left(20+0\right)×5$
$⇒S3=50m$
Thus, the total distance covered by the truck is $S=S1+S2+S3=100+400+50=550m$
Step 6: Determine the average speed of the truck
$Averagespeed=\frac{Totaldistance}{Totaltimetaken}$
$=\frac{S}{t}=\frac{550}{35}=15.7m{s}^{-1}$
(a) The total time for which the truck is in motion is $35s$.
(b) The average speed of the truck for its entire motion is $15.7m{s}^{-1}$.