A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and an 18-wheel truck and accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24.0 m/sm/s when it reaches the end of the ramp, which has length 110 mm . Required:
a. What is the acceleration of the car?
b. How much time does it take the car to travel the length of the ramp?
c. The traffic on the freeway is moving at a constant speed of 23.0 . What distance does the traffic travel while the car is moving the length of the ramp?

a. Acceleration of the car is 2.62 m/s²

b. Time it takes the car to travel the length of the ramp is 9.16 secs

c. Distance traveled by the traffic is 210.7 m

Explanation:

(a) To calculate the acceleration of the car,

From the question,

The car starts from rest, that is

The initial velocity, u = 0 m/s

Final velocity, v = 24.0 m/s

Distance covered by the car = 110 m

acceleration, a = ??

From one of the equations of kinematics for linear motion,

v² = u² +2as

Where v is the final velocity

u is the initial velocity

a is the acceleration

and s is the distance

Hence,

24.0² = 0² + 2(a)(110)

576 = 0 + 220a

576 = 220a

a = 576/220

a = 2.62 m/s²

This is the acceleration of the car

(b) For the time it takes the car to travel the length of the ramp

From

v = u + at

where t is the time

Then,

24.0 = 0 +2.62(t)

2.62t = 24.0

∴ t = 24.0/2.62

t = 9.16 secs

This is the time it takes the car to travel the length of the ramp

(c) For the distance the traffic travel while the car is moving the length of the ramp,

Velocity of the traffic = 23.0 m/s

Time spent by the traffic while the car is moving the length of the ramp = Time spent by the car while moving the length of the ramp

That is,

Time spent by the traffic while the car is moving the length of the ramp = 9.16 secs

From,

Velocity = DIstance / Time

Distance = Velocity × Time

Disatnce = 23.0 × 9.16

Distance = 210.68 m

Distance ≅ 210.7 m

This is the distance traveled by the traffic

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

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from wikipedia