Question

A) A car accelerates uniformly at 3.7 m s⁻² as it passes a stationary road train. The initial speed of the car is 30 m s⁻¹, and it reaches the speed limit of 130 km h⁻¹ as it passes the front of the cab. Calculate the length of the road train. Give your answer to 3 significant figures. include units in your answer.
b) The braking distance of a road train travelling at 15 m s⁻¹ is 70 m. Assuming that the same deceleration is applied at all speeds, calculate the braking distance of a road train travelling at 25 m s⁻¹

Answer 1

(a) The length of the train is 54.6 m

(b)  The braking distance of the road train is 194.44 m

The given parameters:

acceleration of the car, a = 3.7 m/s²

initial velocity of the car, u = 30 m/s

final velocity of the car, v = 130 km/h = 36.11 m/s

To find:

• the length of the train

The length of the train is the distance travelled by the car

The distance traveled by the car is calculated as:

$v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\s = \frac{ v^2 - u^2}{2a} \\\\where;\\\\s \ is \ the \ distance \ travelled \ by \ the \ car\\\\s = \frac{(36.11)^2 - (30)^2}{2\times 3.7} \\\\s = 54.585 \ m\\\\s \approx 54.6 \ m$

Thus, the length of the train is 54.6 m

(b) The braking distance of a road train travelling at 25 m s⁻¹

$a = \frac{v_1^2}{2s_1} = \frac{v_2^2}{2s_2} \\\\Given;\\\\s_1 = 70 \ m\\\\v_1 = 15 \ m/s\\\\v_2 = 25 \ m/s\\\\s_2 = ?\\\\2s_2v_1^2 = 2s_1v_2^2\\\\s_2 = \frac{ 2s_1v_2^2}{2v_1^2} \\\\s_2 = \frac{s_1v_2^2}{v_1^2} \\\\s_2 = \frac{70\times 25^2}{15^2} \\\\s_2 = 194.44 \ m$

Thus, the braking distance of the road train is 194.44 m

Learn more here: brainly.com/question/19572178