A stationary car A is passed by a car B moving with a constant velocity of 15m/s. Two seconds later, car A starts moving with a constant acceleration of 1m/s^2 in the same direction.(If car has constant velocity acceleration must be zero)
a) Using the formula s=ut + 1/2at^2,where s is the distance from the starting point in meters, u is the velocity initially, a is acceleration in m/s^2 and t is the time in seconds, write a relationship between s and t for the car A and car B.
b) Sketch graphs of both functions, on the same set of axes, placing s on the vertical and t on the horizontal.
c) Find the value of t which satisfies both of these equations.
d) What does this value of t mean in relation to the position of the two cars relative to each other.© BrainMass Inc. brainmass.com June 19, 2018, 6:25 am ad1c9bdddf
a) Firstly since car B is moving with constant velocity, its acceleration =0. We can write the relationship between s and t as below:
Car B: sB=15*t (m)
For car A, initial velocity =0 (u=0) ...
The solution provides a detailed and step-by-step explanation (incl. graph) for the problem.