Question

eC https:/session.masteringphysics.com/my KHomework Assignment 1 (Chapter 2) Problem C2-06 Constants A car travels north at 29.700 m/s. The driver applies the brakes and slows the car to 14.000 m/s for 140.40 m Assume the motion is at uniform acceleration.

1)What is the magnitude of the acceleration in m/s^2?

2)What is the direction of the acceleration? (N,S,E,W).

3)If the brakes are applied until the care comes tia stop, what is the total distance covered?

4) how much time elapsed between the instant when the brakes were applied and when the car had traveled a distance of d/2?

Answer 1

Steps

1) use equation of motion

v^2 = u^2 + 2*a*s , to get accelaration

2 ) direction of retardation accelaration is always opposite to the direction of motion

3) Again use equation of step 1 to get total distance covered

4) Again use ist step equation to get velocity at a distance d/2

Now use equation v= u + a* t

to get time elapsed .
Steps 1) use equation of motion v^2 = u^2 + 2*a*s, to get acceleration 2) direction of retardation acceleration is always opposite to the direction of motion 3) Again use equation of step 1 to get total distance covered 4) Again use is t step equation to get velocity at a distance d/2 Now use equation v = u + a* t to get time elapsed . Given: Initial velocity u = 29.7 m/s velocity, after break is applied v = 14.0 m/s distance covered s = 140.4 m PART-1 Now using equation of motion v^2 = u^2 + 2 as rightarrow 14^2 = 29.7^2 + 2a times 140.4 a = -2.4433 m/s^2 P{minus shows acceleration is retardation Hence magnitude of acceleration = 2.4433 m/s^2 \r\n PART - 2 Since the car is travelling in north direction, so the acceleration is south (s) {As the acceleration is retarding in nature and acts opposite to the direction of motion} PART - 3 Now final velocity v = 0 and acceleration is const a = -2.443 m/s^2