Question

If a 1300-kg car can accelerate from 35 km????h to 65 km????h in 3.8 s, how long will it take to accelerate from 55 km????h to 95 km????h? Assume the power stays the same, and neglect frictional losses.

Answer 1

To find the time it takes for the car to accelerate from 55 km/h to 95 km/h, we can use the formula for acceleration. Converting the velocities to m/s and using the given parameters, we find that it would take approximately 3.77 seconds.

Step-by-step explanation:

To solve this problem, we can use the formula to calculate acceleration:

a = (v - u) / t

where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time. We can rearrange this formula to solve for time:

t = (v - u) / a

In the first scenario, the car accelerates from 35 km/h to 65 km/h in 3.8 s. Converting the velocities to m/s gives us:

u = (35 km/h) imes (1000 m/km) / (3600 s/h) = 9.72 m/s

v = (65 km/h) imes (1000 m/km) / (3600 s/h) = 18.06 m/s

a = (18.06 m/s - 9.72 m/s) / 3.8 s = 2.19 m/s^2

Using this information, we can find the time required to accelerate from 55 km/h to 95 km/h:

t = (18.06 m/s - 9.72 m/s) / 2.19 m/s^2 = 3.77 s

Answer 2

6.76 s

Step-by-step explanation:

Power = Energy/time

P = E/t ....................... Equation 1

Where P = Power, E = kinetic energy, t = time.

But,

E = 1/2m(Δv)²................... Equation 2

Where m = mass of the car, Δv = change in velocity.

Substitute equation 2 into equation 1

P = 1/2m(Δv)²/t................... Equation 3

Given: m = 1300 kg, Δv = 65-35 = 30 km/h = 30(1000/3600) m/s = 8.33 m/s, t = 3.8 s

Substitute into equation 3

P = 1/2(1300)(8.33²)/3.8

P = 11869.15 W.

Assuming the power stays the same,

making t the subject of formula in equation 3

t = 1/2m(Δv)²/P.................... Equation 4

Given: m = 1300 kg, Δv = 95-55 = 40 km/h = 11.11 m/s, P = 11869.15 W

Substitute into equation 4

t = 1/2(1300)(11.11²)/11869.15

t = 6.76 s