Solve the Golf Cart Physics Problem & Find Total Trip Time | Physics 101 HW Help

[spoonthrower]

A golfer rides in a golf cart at a speed of 3.10 m/s for 21.0 s. She then gets out of the cart and starts walking at an average speed of 1.20 m/s. for how long (in sec) must she walk if her average speed for the entire trip, riding and walking, is 1.90 m/s?

I know average speed = d/t I'm just confused b/c the distance and the time for the walking part is not given. Please help. Thanks.

This HW problem is for my college physics 101 class and nobody has any ideas? c'mon

 

[Rozenwyn]

Just set up an equation.
[tex] 3.1 \times 21 + 1.2 \times x = 1.9 \times (21+x)[/tex], i.e. The total distance traveled by traveling by the cart and by walking must be equal to the total hypothetical distance traveled if he "travelled" at a speed of 1.9 m/s on whatever.
Solve for x.

 

[kyle8921]

Hello,

I can help you solve this golf cart physics problem. First, let's write out the information we have:

Initial speed of golf cart (v1) = 3.10 m/s
Time spent riding in golf cart (t1) = 21.0 s
Average speed for entire trip (v2) = 1.90 m/s
Average speed while walking (v3) = 1.20 m/s

To find the total trip time (t2), we can use the equation for average speed:

v2 = (d1 + d2) / (t1 + t2)

Where d1 is the distance travelled in the golf cart and d2 is the distance travelled while walking.

We can also use the equation for average speed to find the distance travelled while walking:

v3 = d2 / t2

Now, we can rearrange the equations to solve for t2:

t2 = (d2 / v3) = (v2 * t1) - (d1 / v1)

Substituting in the values we have, we get:

t2 = (v2 * t1) - (d1 / v1)
= (1.90 * 21.0) - (d1 / 3.10)

To find d1, we can use the equation d = v * t:

d1 = v1 * t1
= 3.10 * 21.0
= 65.10 m

Substituting this into our previous equation, we get:

t2 = (1.90 * 21.0) - (65.10 / 3.10)
= 39.90 - 21.00
= 18.90 s

Therefore, the golfer must walk for 18.90 seconds to have an average speed of 1.90 m/s for the entire trip. I hope this helps! Good luck with your physics homework.