A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

Question

3 months ago

13

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

are both traveling at the same speed of 18.0 m/s, and the distance between them is 58.0 m . After t1 = 5.00 secs, the motorcycle starts to accelerate at a rate of 4.00 m/s^2 . The motorcycle catches up with the car at some time t2 . A) How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find t2-t1 Express the time numerically in seconds using three significant figures. B) How far does the motorcycle travel from the moment it starts to accelerate (at time t1 ) until it catches up with the car (at time t2)? Should you need to use an answer from a previous part, make sure you use the unrounded value. Answer numerically in meters using three significant figures.

Physics

1 answer:

Softa [3K] · Answer 1 · 2026-02-21T05:37:49+03:00

Answer:

a) $\Delta{t} = 5.39s$

b) the distance the motorcycle covers is 155 m

Explanation:

Let $t_2-t_1 = \Delta{t}$ denote the variables. Next, we analyze the motion equation for the accelerating motorcycle alongside the constant speed of the car:

$v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}$

where:

$v_{m2}$ represents the motorcycle's speed at time 2

$v_{c}$ is the steady velocity of the car

$v_{0}$ indicates the initial speeds of both vehicle types at time 1

d signifies the distance separating the car and motorcycle at the initial moment

x is the distance the car travels from time 1 to time 2

Solving the equations provides:

$\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]$

$v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s$

For the second query, we determine x+d by applying the car’s motion equation to compute x:

$x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933$