What is the speed of the sports car at the second marker?

[miamiheat5]

A sports car, picking up speed, passes between two markers in a time of 4.1s. The markers are separated by 120m. All the while, the car accelerates at 1.8m/s^2. What is its speed at the second marker?

Here is what i did:

d = vi(t)+1/2(a)t^2
120=vi(4.1)+1/2(1.8)(4.1)^2
29.3=vi+15.1
vi=14.2m/s

d=1/2(vi+vf)t
120=1/2(14.2+vf)(4.1)
vf=44.3m/s
vf=25.2m/s

 

[radou]

A sports car, picking up speed, passes between two markers in a time of 4.1s. The markers are separated by 120m. All the while, the car accelerates at 1.8m/s^2. What is its speed at the second marker?

Here is what i did:

d = vi(t)+1/2(a)t^2
120=vi(4.1)+1/2(1.8)(4.1)^2
29.3=vi+15.1
vi=14.2m/s

d=1/2(vi+vf)t
120=1/2(14.2+vf)(4.1)
vf=44.3m/s
vf=25.2m/s

Your calculation is wrong. Check the numbers once again.

 

[m2003]

Based on the given information, the speed of the sports car at the second marker would be approximately 25.2m/s. This is calculated by using the equation d=1/2(vi+vf)t, where d is the distance between the two markers (120m), vi is the initial velocity (14.2m/s), vf is the final velocity (which we are trying to find), and t is the time it takes for the car to travel between the markers (4.1s). We also use the equation d = vi(t)+1/2(a)t^2 to find the initial velocity, which is then used in the first equation to solve for vf. The acceleration of the car (1.8m/s^2) is also taken into account in both equations. Therefore, the speed of the sports car at the second marker is 25.2m/s.