Physics, 22.09.2019 20:50 ana2896

Assume the truck is going at v0 = 25 m/s, and you're 20 m behind the truck, initially going the same velocity. you decide to accelerate at a = 1.0 m/s2 to pass the truck. to pass safely, you also want to pass the truck by 20 m before moving back to the lane. in part b, you will be asked to find the time required to pass (assuming that you accelerate at the same rate the entire time).

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Answers

The time required to pass the truck from distance 20 m behind the truck to distance 20 m in front of the truck is about 8.9 s

Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem !

Given:

vo = 25 m/s

a = 1.0 m/s²

Unknown:

t = ?

Solution:

The time needed for the car from the distance of 20 m behind the truck to a distance of 20 m in front of the truck is:

$d_{truck} = d_{car} - 20 - 20$

$v_o \times t = v_o \times t + \frac{1}{2} \times a \times t^2 - 40$

$25t = 25t + \frac{1}{2} \times 1 \times t^2 - 40$

$\frac{1}{2}t^2 = 40$

t^2 = 80

$t = \sqrt{80}$

$t = 4\sqrt{5} ~ s$

$\boxed {t \approx 8.9 ~ s}$

Learn moreVelocity of Runner : Kinetic Energy : Acceleration : The Speed of Car : Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

Assume the truck is going at v0 = 25 m/s, and you're 20 m behind the truck, initially going the same

Answer from: slugmilk1090

The time required to come in the same lane as of the truck is $\boxed{{\text{8}}{\text{.94 s}}}$ .

Further explain:

We have to calculate the time required to come in the same lane as of the truck.

Given:

Velocity of the truck is $25{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

Initial velocity of mine is $25{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

Initial distance between the truck and the observer (me) is $20{\text{ m}}$ .

Uniform acceleration of the observer is $1{{{\text{ m}}} \mathord{\left/ {\vphantom {{{\text{ m}}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}$ .

Safe distance between truck and observer before moving back to the lane is $20{\text{ m}}$ .

Formula and concept used:

The expression for the distance travelled by the truck is given as.

${s_{{\text{truck}}}} = vt$ …… (1)

Here, is the speed of truck and is the time.

Now, the time taken by the observer to come at same lane can be calculated by the kinematic equation of motion.

$\boxed{s=ut+\frac{1}{2}a{t^2}}$ …… (2)

Here, is the initial distance between truck and the observer, is the acceleration of the observer and is the time.

Calculation:

Substitute $25\,{\text{m/s}}$ for in equation (1).

${s_{{\text{truck}}}} = \left( {25\,{\text{m/s}}} \right)t$

Now, the distance traveled by observer with respect to the truck while the observer comes back to the lane is,

$\begin{gathered}s=20+20\\=40{\text{ m}}\\\end{gathered}$

The expression for the total distance is given as.

$s={s_{{\text{truck}}}}+40\,{\text{m}}$

Substitute $\left( {25\,{\text{m/s}}} \right)t$ for ${s_{{\text{truck}}}}$ in above expression.

$s=\left( {25\,{\text{m/s}}} \right)t+40\,{\text{m}}$

Substitute $\left( {25\,{\text{m/s}}} \right)t + 40\,{\text{m}}$ for , $25\,{\text{m/s}}$ for and $1\,{\text{m/}}{{\text{s}}^{\text{2}}}$ for in equation (2).

$\left( {25\,{\text{m/s}}} \right)t + 40\,{\text{m}}=\left( {25\,{\text{m/s}}} \right)t + \dfrac{1}{2}\left( {1\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right){t^2}$

Rearrange the above expression we get.

$\begin{aligned}t&=\sqrt {80\,{\text{s}}}\\&=\,{\text{8}}{\text{.94}}\,{\text{s}}\\\end{aligned}$

Thus, the time required for car to come in same lane is $8.94\,{\text{s}}$ .

Learn more:

1. Component of the lever: .

2. Type of mirror used by dentist: .

3. Motion under force .

Answer detail:

Grade: Senior School

Subject: Physics

Chapter: Kinematics

Keywords:

The truck, you are behind the truck, same velocity, pass the truck, moving back to the lane, time required to pass, same acceleration, overtake, motion, velocity, distance covered, 8.94s, 8.9s, equation of motion.

Answer from: Quest

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Answer from: Quest

access to city

explanation:

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