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?

(a) For how much time the truck is in motion?

(b) What is the average speed of the truck for its entire motion?

meakQueueksiu7e

Beginner2023-03-26Added 5 answers

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 $v=u+at$

$\Rightarrow t1=\frac{v-u}{a}$

$\Rightarrow t1=\frac{20-0}{2}$

$\Rightarrow t1=10s$

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 $S1$

$\Rightarrow S1=0+\frac{1}{2}\times 2\times {\left(10\right)}^{2}$

$\Rightarrow S1=100m$

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,

$\Rightarrow S2=20\times 20+0$

$\Rightarrow S2=400m$

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, $S=\frac{1}{2}\left(v+u\right)t$

$\Rightarrow S3=\frac{1}{2}\left(20+0\right)\times 5$

$\Rightarrow S3=50m$

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

$Averagespeed=\frac{Totaldistance}{Totaltimetaken}$

$=\frac{S}{t}=\frac{550}{35}=15.7m{s}^{-1}$

Final Answer:

(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 $15.7m{s}^{-1}$.

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 $v=u+at$

$\Rightarrow t1=\frac{v-u}{a}$

$\Rightarrow t1=\frac{20-0}{2}$

$\Rightarrow t1=10s$

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 $S1$

$\Rightarrow S1=0+\frac{1}{2}\times 2\times {\left(10\right)}^{2}$

$\Rightarrow S1=100m$

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,

$\Rightarrow S2=20\times 20+0$

$\Rightarrow S2=400m$

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, $S=\frac{1}{2}\left(v+u\right)t$

$\Rightarrow S3=\frac{1}{2}\left(20+0\right)\times 5$

$\Rightarrow S3=50m$

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

$Averagespeed=\frac{Totaldistance}{Totaltimetaken}$

$=\frac{S}{t}=\frac{550}{35}=15.7m{s}^{-1}$

Final Answer:

(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 $15.7m{s}^{-1}$.

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