Question

A fugitive tries to hop on a freight train traveling at a constant speed of 5.0 m/s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 3.8 m/s2 to his maximum speed of 8.0 m/s.

Answer 1

Let d =  distance that the fugitive travels to get on the train.

Let t =  the time to travel the distance d.
The fugitive starts from rest accelerates at a = 3.8 m/s².
Therefore
(1/2)*(3.8 m/s²)*(t s)² = (d m)
1.9 t² = d                      (1)

The train travels at constant speed  5.0 m/s.
Therefore
(5.0 m/s)*(t s) = d    
5t = d                          (2)

If the fugitive successfully boards the train, then equate (1) and (2).
1.9t² = 5t
t = 0 or t = 2.6316 s
Ignore t = 0, so t = 2.6316 s.

The speed of the fugitive after 2.6316 s, is
v = (3.8 m/s²)*(2.6316 s) = 10 s

This speed exceeds the maximum speed of the fugitive, therefore the fugitive fails to get on the train.

Answer: The fugitive fails to get on the train.

Answer 2

A. The time to catch up to the empty box car is about 2.8 s

B. The distance traveled to reach the box car is about 14 m

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:

Acceleration of car = a = 3.8 m/s²

Initial Velocity of car = u = 0 m/s

Velocity of train = vt = 5.0 m/s

Unknown:

A . Time Taken = t = ?

B. Distance = d = ?

Solution:

Question A:

Firstly, we calculate for the distance traveled by the car until it reaches a maximum speed of 8 m / s.

$v^2 = u^2 + 2ad_1$

$8^2 = 0^2 + 2(3.8)d_1$

$64 = 7.6d_1$

$\boxed {d_1 = \frac{160}{19} ~ m}$

Next , we find time taken for the car to reach this maximum speed

$v = u + a t$

$8 = 0 + 3.8t'$

$t' = 8 \div 3.8$

$\boxed {t' = \frac{40}{19} ~ s}$

When trains and cars meet, then

$d_{train} = d_{car}$

$v_t \times t = d_1 + v_{max} \times (t - t')$

$5t = \frac{160}{19} + 8(t - \frac{40}{19})$

$5t = \frac{160}{19} + 8t - \frac{320}{19}$

$5t = -\frac{160}{19} + 8t$

$8t - 5t = \frac{160}{19}$

$3t = \frac{160}{19}$

$t = \frac{160}{19} \div 3$

$t = \frac{160}{57} ~ s$

$\large {\boxed {t \approx 2.8 ~ s} }$

Question B:

$d_{car} = -\frac{160}{19} + 8t$

$d_{car} = -\frac{160}{19} + 8 \times \frac{160}{57}$

$d_{car} = \frac{800}{57} ~ m$

$\large {\boxed {d_{car} \approx 14 ~ m} }$

Learn more

• Velocity of Runner : brainly.com/question/3813437
• Kinetic Energy : brainly.com/question/692781
• Acceleration : brainly.com/question/2283922
• The Speed of Car : brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

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