What is the Velocity of a Frictionless Car Passing Through Two Photo Gates?

[CollegeStudent]

Homework Statement

A frictionless car on a straight track tipped at 8.21° accelerates at 1.40 m/s².
Released from rest and travels through 2 photo gates. First is 18.8cm away from starting position. Second is 14.8 cm from the first one. Whats the velocity as the car passes through each gate?

Homework Equations

I'm not sure here. The angle throws me off since the professor hasn't gone over an equation with an angle involved yet.

The Attempt at a Solution

I just could use the formula that would be used. I tried using Vf² = Vi² + 2aΔx ... but then realized the angle wasn't taken into account

 

[Doc Al]

but then realized the angle wasn't taken into account

The effect of the angle is included in the acceleration, which they gave you. So you do not have to worry about the angle. (Given the angle, you could have deduced the acceleration. But you haven't covered that part yet, so they just told you the answer.)

 

[CollegeStudent]

Oh okay, so basically if I hadn't had the acceleration, I could have figured it out somehow with the angle? okay.

So now what I did was: by using

Vf² = Vi² + 2aΔx

So

Vf² = 0 + 2(1.40m/s/s)(18.8cm)
so
Vf² = 0 + 2(1.40m/s/s)(.188m)
Vf² = 0 + (2.80m/s/s)(.188m)
Vf² = .5264
Vf = .7255
Does this sound right?

 

[Doc Al]

Oh okay, so basically if I hadn't had the acceleration, I could have figured it out somehow with the angle? okay.

Right. You'll get to that when you study dynamics, most likely.

So now what I did was: by using

Vf² = Vi² + 2aΔx

So

Vf² = 0 + 2(1.40m/s/s)(18.8cm)
so
Vf² = 0 + 2(1.40m/s/s)(.188m)
Vf² = 0 + (2.80m/s/s)(.188m)
Vf² = .5264
Vf = .7255
Does this sound right?

Perfect!

 

[CollegeStudent]

Awesome thank you Doc Al !