Question

Two drivers agree to race each other on a straight road. Car A has a top speed of 80 km/h and an acceleration of 5 m/s^2. Car B has a top speed of 95 km/h and an acceleration of 2 m/s^2. Both cars accelerate to their top speeds and maintain their speeds for the whole race. What is the distance between car A and B after 10 minutes?

Answer 1

The 80 and 95 km/h speeds of cars A and B and their 5 and 2 m/s² acceleration gives the distance between them after 10 minutes as approximately 2,375.3 meters

What are the equations of the motion of the cars?

Top speed of car A = 80 km/h

$80 km /h = 22 \frac{2}{9} \: m/ s$

Acceleration of car A = 5 m/s²

Top speed of car B = 95 km/h

$95 \: km /h = 26 \frac{7}{18} \: m/ s$

Acceleration of car B = 2 m/s²

Time of travel = 10 minutes

The time, t1, it takes car A to reach the top speed is found as follows;

• v = u + a•t

Where;

v = Final velocity (speed)

u = Initial velocity (speed)

a = Acceleration

t = Time

Whereby both cars start from rest, we have;

u = 0

v = 0 + a•t = a•t

• t = v/a

Which gives;

$t_1 = \frac{22 \frac{2}{9}}{5} \: s ≈ 4 \frac{4}{9} \: s$

The distance covered while accelerating is given by the formula;

s = u•t + 0.5•a•t²

Which gives;

• s1 = 0.5 × 5 × (4 4/9)²

$s_1 = 0.5 \times 5\times \left(4 \frac{4}{9} \right)^2 ≈ 49 \frac{31}{81}$

$s_1 = 49 \frac{31}{81} \: m$

The time of travel at top speed, t2, is given as follows;

$t_2 = 10 × 60 - 4 \frac{4}{9} = 595 \frac{5}{9}$

Distance, s2 traveled at top speed is therefore;

$s_2 = 22 \frac{2}{9} × 595 \frac{5}{9} = 13243 \frac{46}{81}$

The distance traveled by car A is therefore;

$d_A = 49 \frac{31}{81} + 13243 \frac{46}{81} =13283 \frac{77}{81}$

Similarly, for car B, we have;

$t_1 = \frac{26 \frac{7}{18}}{2} \: s = 13 \frac{7}{36} \: s$

The distance covered while accelerating is given by the formula;

$s_1 = 0.5 \times 2 \times \left(13 \frac{7}{36} \right)^2 = 174 \frac{121}{1296}$

$s_1 =174 \frac{121}{1296} \: m$

The time of travel at top speed, t2, is given as follows;

$t_2 = 10 × 60 - 13 \frac{7}{36} = 586 \frac{29}{36}$

Distance, s2 traveled at top speed is therefore;

$s_2 = 26 \frac{7}{18} × 586 \frac{29}{36} = 15485 \frac{95}{648}$

The distance traveled by car B is therefore;

• $d_B = 174 \frac{121}{1296} + 15485 \frac{95}{648} = 15659 \frac{311}{1296}$

The distance between car A and car B in 10 minutes, is therefore;

• $15659 \frac{311}{1296} - 13283 \frac{77}{81} = 2375\frac{125}{432}$

The distance between car A and car B after 10 minutes is d ≈ 2375.3 m.

Learn more about the equations of motion here:

brainly.com/question/10727726

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