Two cars cover the same distance in a straight line. Car A covers the distance at a constant velocity. Car B starts from rest and maintains a constant acceleration. Both cars cover a distance of 491 m in 220 s. Assume that they are moving in the +x direction. Determine (a) the constant velocity of car A, (b) the final velocity of car B, and (c) the acceleration of car B.

(a)

(b)

(c)

Explanation:

The kinematic equation of car A is:

On the other hand, the kinematic equations of car B are:

First, we can compute the constant velocity of car A from (1):

Next, we can obtain the acceleration of car B from (3):

Finally, using the equation (2), we can get the final velocity of car B:

In words, we got that the constant velocity of car A is 2.23m/s (a), the final velocity of car B is 4.46m/s (b), and the acceleration of car B is 0.0202m/s² (c).

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s