Question

1) Brakes applied to a car produce an acceleration of 5m/s2 in the opposite direction to the motion. If car takes 1.5s to stop after applying brakes,calculate the distance travelled by it.
2) A person walks along the sides of a square field each side is 100m long. What is the maximum magnitude of displacement of the person in any time interval?

Answer 1

1)  you can do this in two steps or in one step.

deceleration by brakes a = -5m/sec²
time duration of travel t = 1.5 sec
V = U + a t       V = 0 as car stopped
0  =  U - 5 *1.5  => U = 7.5 m/sec

Distance traveled S = U t + 1/2 a t²
        = 7.5 * 1.5 - 1/2 5 * 1.5²  = 5.625 meters

You can do this in ONE STEP if you use the formula  S = V t - 1/2 a t²
   S = 0 * t - 1/2 (-5) 1.5²  = 5.625 meters
================================
2)
  See Figure 1  and Figure 2.
Let us say he starts from center of side from A.  DIsplacement vector is vector joining this position to his current position. It can be one of the green lines.  See the maximum of them. It is AD or AE. 

Let us say he starts walking from a corner, say B. Displacement vector at some positions are shown.  Maximum of them is BD.  As BD is more than AD, BD is the largest displacement vector and also, AC = BD.

Magnitude of max displacement vector = BD or AC = √2 * 100 = 141.4 meters

Answer 2

Answer:

S=vt-1/2×at^2

S=0×t-1/2×(-5)1.5^2

=5.625 meters Ans.