The time required to come in the same lane as of the truck is[tex]\boxed{{\text{8}}{\text{.94 s}}}[/tex].
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 [tex]25{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}[/tex].
Initial velocity of mine is [tex]25{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}[/tex].
Initial distance between the truck and the observer (me) is [tex]20{\text{ m}}[/tex].
Uniform acceleration of the observer is [tex]1{{{\text{ m}}} \mathord{\left/ {\vphantom {{{\text{ m}}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}[/tex].
Safe distance between truck and observer before moving back to the lane is [tex]20{\text{ m}}[/tex].
Formula and concept used:
The expression for the distance travelled by the truck is given as.
[tex]{s_{{\text{truck}}}} = vt[/tex] …… (1)
Here, [tex]v[/tex] is the speed of truck and [tex]t[/tex] is the time.
Now, the time taken by the observer to come at same lane can be calculated by the kinematic equation of motion.
[tex]\boxed{s=ut+\frac{1}{2}a{t^2}}[/tex] …… (2)
Here, [tex]u[/tex] is the initial distance between truck and the observer, [tex]a[/tex] is the acceleration of the observer and [tex]t[/tex] is the time.
Calculation:
Substitute [tex]25\,{\text{m/s}}[/tex] for [tex]v[/tex] in equation (1).
[tex]{s_{{\text{truck}}}} = \left( {25\,{\text{m/s}}} \right)t[/tex]
Now, the distance traveled by observer with respect to the truck while the observer comes back to the lane is,
[tex]\begin{gathered}s=20+20\\=40{\text{ m}}\\\end{gathered}[/tex]
The expression for the total distance is given as.
[tex]s={s_{{\text{truck}}}}+40\,{\text{m}}[/tex]
Substitute [tex]\left( {25\,{\text{m/s}}} \right)t[/tex] for [tex]{s_{{\text{truck}}}}[/tex] in above expression.
[tex]s=\left( {25\,{\text{m/s}}} \right)t+40\,{\text{m}}[/tex]
Substitute [tex]\left( {25\,{\text{m/s}}} \right)t + 40\,{\text{m}}[/tex] for [tex]s[/tex], [tex]25\,{\text{m/s}}[/tex] for [tex]v[/tex] and [tex]1\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex] for [tex]a[/tex] in equation (2).
[tex]\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}[/tex]
Rearrange the above expression we get.
[tex]\begin{aligned}t&=\sqrt {80\,{\text{s}}}\\&=\,{\text{8}}{\text{.94}}\,{\text{s}}\\\end{aligned}[/tex]
Thus, the time required for car to come in same lane is [tex]8.94\,{\text{s}}[/tex].
Learn more:
1. Component of the lever: https://brainly.com/question/1073452.
2. Type of mirror used by dentist: https://brainly.com/question/997618
.
3. Motion under force https://brainly.com/question/6125929.
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.
