Question

A driver operating a vehicle at a speed of 25.0 m/s intends to pass a second vehicle 2.00 km ahead traveling in the same direction with a speed of 30.0 m/s. The driver of the first vehicle accelerates at a uniform rate to pass the second vehicle, and does so after traveling 4.00 km. Thus, whereas the first (rear) vehicle traverses 4.00 km, the second (front) car only travels 2.00 km, when it is passed. If the driver of the second vehicle, not wishing to be passed, begins to accelerate at a rate of 0.400 m/s^2 the moment the first car begins its own acceleration, how much time elapses before the rearward (i.e. first) car passes the vehicle in the front (i.e. second car)? 35.0 s 50.0 s 65.0 s 75.0 s #15 Road Warriors (continued) In #14. what is the uniform acceleration of the rearward (i.e. first) vehicle? 0.800 m/s^2 1.40 m/s^2 2.20 m/s^2 3.40 m/s^2

Answer 1

14) time in the first case = distance travelled by second vehicle / its speed

= 2km/(30m/s) = 66.67s

using second equation of motion for first vehicle,

s = ut + 0.5at^2

4000 = 25*66.67 + 0.5*a*66.67^2

a = 2333/[0.5*66.67^2]

= 1.05 m/s^2

now in second case, let they both meet at t

distance travelled by first vehical = distance travelled by first vehical + 2000

25t + 0.5*1.05 t^2 = 30t + 0.5*0.4*t^2 +2000

(0.5*1.05-0.5*0.4) t^2 - 5t -2000 = 0

0.325 t^2 -5t -2000 =0

solving the quadratic equation,

t =86 s, and a=1.05 should be correct answer of 14 and 15 respectively [checked many times]

However, among the given options, following is correct

Answer of Q14= B, 50s

Answer of Q15 = C 2.2m/s^2