Question

a) A car is driven along a curve at a speed of 38 m/s. If the car

Answer 1

a)

normal acceleration, An = 11.2 m/s2

But, An = v^2/R

So, R = v^2/An = 38^2/11.2 = 128.9 m <----------answer

b)

displacement = 2*r = 1.6*2 = 3.2 km <--------answer

Explanation:

For 2.5 laps we abserved he would complete 2 full circles and then half circle. Thus he would be finally at the opposite end of the circle after 2.5 laps.

So, he would be at a distance equal to the diameter = 2*radius

For distance , total distance = 2.5 times the circumference = 2.5*(2*pi*r) = 5*pi*r

= 5*pi*1.6 = 25.13 m <-------answer

c)

Using the equation of motion,

v = u +at

u = initial speed = 0 m/s

a = 9.8 m/s2

after 3s, speed of the 1st jumper, v = 0+9.8*3 = 29.4 m/s

So, after 3 s, now their speeds at t=3 are 29.4 m/s (first diver) and 0(second diver. As he just dived now)

So, relative speed now, u,rel= 29.4 - 0 29.4 m/s

relative acceleration, a,rel = 0 as both have same acceleration

So,after t = 4s,

v,rel = u,rel + a,rel*t

So, v,rel = 29.4 + 0*4 = 29.4 m/s <------answer