Applied Maths LC Paper 1991 Q1 Linear Motion HELP!!!!

leavingcerttt

hi could someone please explain how to do this question?

1991
(a) A particle starts from rest at a point p and accelerates at 2m/s2 until it reaches a speed v m/s. It travels at this speed for 1 minute before decelerating at 1 m/s2 to rest at q. The total time for the journey is 2 minutes.
(i) Calculate the distance pq. (ans=3600m)
(ii) If a second particle starts from p at time t = 0 and moves along pq with speed (2t + 50)m/s, find the time taken to reach q. (ans=40secs)

THANK YOU!

Raspberry Fileds

Struggling with both parts?

Raspberry Fileds

t1 + t3 = 60
a = acceleration = 2
b = deceleration = 1

Tan A = a = v/t1
Tab B = b = v/t3

v/2 + v/1 = t1 + t3
3v/2 = 60
v = 40

1/2(t1)(v) + 1/2(t3)(v) + (t2)(v) = d
1/2(t1+t3)(v) + (t2)(v) = d
1/2(60)(40) + (60)(40) = d = 3600


ii) The mistake many will have made is assuming the second particle started from rest.

At t=0, v = 2(0) + 50 = 50.

The coefficient of t gives the acceleration - i.e. 2.

u = 50
v = 2t + 50
a = 2
s = 3600
t = t

3600 = 50t + 1/2(2)(t^2)
(t^2) + 50t - 3600 = 0

The quadratic formula will give you two roots - one of which will be negative. As time can only increase, and the particle reaches q after t=0, the positive root is the answer.