Question

Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and accelerates at 15.0 km/h2 for 20.0 min. Car C starts out traveling at 32.0 km/h and accelerates at 40.0 km/h2 for 30.0 min. Car D starts out traveling at 110.0 km/h and decelerates at 60.0 km/h2 for 45.0 min. Which car's final speed was closest to 15 m/s? A. A B. B C. C D. D

Answer 1

To determine which car's final speed was closest to 15 m/s, we need to calculate the final speed for each car. Comparing the final velocities, Car A's final speed of 41.25 km/h is closest to 15 m/s.

Step-by-step explanation:

To determine which car's final speed was closest to 15 m/s, we need to calculate the final speed for each car. Car A has an initial speed of 35.0 km/h and accelerates at 25.0 km/h² for 15.0 minutes. Using the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time, we can calculate vf for Car A. Similarly, we can calculate the final velocities for Cars B, C, and D, and compare them to see which one is closest to 15 m/s.

Car A: vf = 35.0 km/h + (25.0 km/h²)(15.0 min) = 35.0 km/h + (25.0 km/h²)(0.25 h) = 35.0 km/h + 6.25 km/h = 41.25 km/h.

Car B: vf = 45.0 km/h + (15.0 km/h²)(20.0 min) = 45.0 km/h + (15.0 km/h²)(0.333 h) = 45.0 km/h + 4.995 km/h = 49.995 km/h.

Car C: vf = 32.0 km/h + (40.0 km/h²)(30.0 min) = 32.0 km/h + (40.0 km/h²)(0.5 h) = 32.0 km/h + 40.0 km/h = 72.0 km/h.

Car D: vf = 110.0 km/h - (60.0 km/h²)(45.0 min) = 110.0 km/h - (60.0 km/h²)(0.75 h) = 110.0 km/h - 33.75 km/h = 76.25 km/h.

Comparing the final velocities for each car, we can see that Car A's final speed of 41.25 km/h is closest to 15 m/s. Therefore, the answer is Car A.

Answer 2

1 kilometre=1000 metre

1 hour = 3600 second

`1\ km/hr=(1000)/(3600) m/s`

`1\ km/hr=(5)/(18) m/s`

The initial velocity of car A is 35.0 km/hr i.e

`35.0\ km/hr=35*(5)/(18) m/s`

= 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as
`25\ km/hr^2`

`=\ 25*(1000)/(3600*3600) m/s^2`

`=0.00192901234 m/s^2`

The time taken by car A = 15 min.

From equation of kinematics we know that-

v= u+at [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A, v = 9.72 m/s +[0.00192901234×15×60]m/s

=11.456111106 m/s

The acceleration of B is given as
`15\ km/hr^2`

`=0.00115740740740 m/s^2`

The time taken by car B =20 min

The final velocity of B is -

v= u+at

= u-at [Here a is negative due to deceleration]

=12.5 m/s +[0.0011574074074×20×60]

=13.8888888.....

=13.9

The acceleration of C is given as
`40\ km/hr^2`

`=\ 0.003086419753 m/s^2`

The time taken by car C =30 min

The final velocity of C is-

v = u+at

=8.89 m/s+[0.003086419753×30×60] m/s

=14.4455555555..m/s

=14.45 m/s

The car C is decelerating.The deceleration is given as-
`60\ km/hr^2`

`=0.0046296296296m/s^2`

The time taken by car D= 45 min.

The final velocity of the car D is-

v =u+at

=30.56 -[0.00462962962962×45×60]m/s

=18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.