Question

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 25.0 m/s when it reaches the end of the ramp, which has length 111 m.1. What is the acceleration of the car?2. How much time does it take the car to travel the length of the ramp?3. 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?

Answer 1

Answer:

1

  $a = 2.82 \ m/s^2$

2

 $t = 8.87 \ s$

3

 $s = 204 \ m$

Explanation:

From the question we are told that

   The  speed of the car is  $v = 25.0 \ m/s$

   The length of the ramp is  $d = 111 \ m$

   The  constant velocity of the traffic $v_t = 23.0 \ m/s$

Generally the acceleration of the car is mathematically represented as

           $a = \frac{v^2 - u^2 }{2d}$

Here  u is  equal to zero given that the car started from rest so

           $a = \frac{25^2 - 0^2 }{2 * 111}$

       =>    $a = 2.82 \ m/s^2$

Generally the time taken is mathematically represented as

      $t = \frac{ v - u}{ a}$

=>   $t = \frac{ 25 - 0}{2.82}$

=>   $t = 8.87 \ s$

The  distance traveled by the traffic is mathematically represented as

  $s = v_{t}t + \frac{1}{2} a t^2$

Here  a is zero given that the traffic was moving at constant speed

=>    $s = 23 * 8.87$

=>     $s = 204 \ m$