A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is 2.5 km away

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

The average time it took the car to travel 0.25 meters was 2.23 seconds

The average time it took the car to travel 0.50 meters was 3.13 seconds

Happy to help :)