Physics help- tangential acceleration

[monaluma]

A 1500kg car starts from rest and drives around a flat 50m-diameter circular track. The forward force provided by the car's drive wheels is a constant 1300N. What is the magnitude of the car's acceleration at t=11s?

 

[dipstik]

you should go to the homework section and state your thoughts on the problem, list the equations you think you need, show whatever wrk you have done so far and explain where you are getting stuck.

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[nlightner]

Based on the information provided, we can use the equation for tangential acceleration, a_t = r * α, where r is the radius of the circular track and α is the angular acceleration. Since the car is driving at a constant speed, we can assume that the angular acceleration is also constant.

First, we need to find the angular velocity of the car. This can be calculated using the formula ω = v/r, where v is the linear velocity of the car. Since the car is starting from rest, its initial velocity is 0. Therefore, ω = 0/r = 0.

Next, we can use the formula α = (ω_f - ω_i)/t, where ω_f is the final angular velocity, ω_i is the initial angular velocity, and t is the time. In this case, the final angular velocity can be calculated using the formula ω_f = 2π/t, where t is the time it takes for the car to complete one full revolution around the track. Since the track has a diameter of 50m, the circumference is 2πr = 2π*25m = 50πm. Therefore, the time it takes for the car to complete one full revolution is t = distance/velocity = 50πm/1300N = 38.37s.

Now, we can calculate the angular acceleration: α = (2π/t - 0)/11s = 0.18rad/s^2.

Finally, we can calculate the tangential acceleration: a_t = r * α = 25m * 0.18rad/s^2 = 4.5m/s^2.

Therefore, the magnitude of the car's acceleration at t=11s is 4.5m/s^2. It is important to note that this is only the tangential acceleration; the car will also experience centripetal acceleration towards the center of the circular track.