Question

5. A walker travels along a straight road passing through the points A and B on the road with speeds 0.9 ms ¹ and 1.3 ms ¹ respectively. The walker's acceleration between A and B is constant and equal to 0.004 ms-².

(i) Find the time taken by the walker to travel from A to B, and find the distance AB. A cyclist leaves A at the same instant as the walker. She starts from rest and travels along the straight road, passing through B at the same instant as the walker. At time t s after leaving A, the cyclist's speed is kt³ ms¯¹, where k is a constant.

(ii) Show that when t = 64.05, the speed of the walker and the speed of the cyclist are the same, correct to 3 significant figures.

(iii) Find the cyclist's acceleration at the instant she passes through B.​

Answer 1

The time taken by the walker to travel from point A to point B is 100 seconds. The distance AB is 170 meters. When t = 64.05 seconds, the speed of the walker and the speed of the cyclist are the same. The cyclist's acceleration at the instant she passes through point B is 9.3805 x 10^-5 m/s².

Step-by-step explanation:

To find the time taken by the walker to travel from point A to point B, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken. In this case, the walker's initial velocity is 0.9 m/s and the final velocity is 1.3 m/s. The acceleration is given as 0.004 m/s². Plugging these values into the equation, we get:

1.3 = 0.9 + (0.004t)

Simplifying the equation, we have:

0.4 = 0.004t

Dividing both sides by 0.004, we get:

t = 100 seconds

So, the time taken by the walker to travel from A to B is 100 seconds.

To find the distance AB, we can use the equation s = ut + 0.5at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time taken. In this case, the initial velocity is 0.9 m/s, the acceleration is 0.004 m/s², and the time taken is 100 seconds. Plugging these values into the equation, we get:

s = (0.9)(100) + 0.5(0.004)(100^2)

Simplifying the equation, we have:

s = 90 + 0.5(0.004)(10000)

s = 90 + 2(40)

s = 90 + 80

s = 170

So, the distance AB is 170 meters.

To show that the speeds of the walker and the cyclist are the same when t = 64.05 seconds, we can substitute the given values into the equations for their speeds. The speed of the walker is given as 1.3 m/s, and the speed of the cyclist is given as kt^3 m/s, where k is a constant. Plugging in t = 64.05 seconds, we get:

1.3 = k(64.05)^3

To find the constant k, we can rearrange the equation:

k = 1.3 / (64.05)^3

Plugging in the values, we get:

k = 6.1303 x 10^-9

So, when t = 64.05 seconds, the speed of the walker and the speed of the cyclist are the same, correct to 3 significant figures.

To find the cyclist's acceleration at the instant she passes through point B, we can differentiate the equation for the cyclist's speed with respect to time. The equation for the cyclist's speed is kt^3 m/s, where k is the constant we found earlier. Differentiating this equation, we get:

Acceleration = d(kt^3)/dt

Acceleration = 3k(t^2)

Plugging in t = 64.05 seconds and k = 6.1303 x 10^-9, we get:

Acceleration = 3(6.1303 x 10^-9)(64.05^2)

So, the cyclist's acceleration at the instant she passes through point B is 9.3805 x 10^-5 m/s².