At any instant t = 0 a motorbike start from rest in a given direction,

1.	At any instant t = 0 a motorbike start from rest in a given direction, a car moving with constant speed overtakes themotorbike at that instant when it is moving with a speed 40 m/s. Motor bike accelerates uniformly till t = 18s andthen move with constant speed and overtakes the car at t = 27 s.The maximum speed of motorbike in m/s isA) 80(B) 60(C) 100(D) 120plz tell
Answer» Answer: 60 m/sec Explanation: At t=0,u=0for the MOTORBIKE And at t=18, v=18a(which will be the Vmax) So, S1=1/2a18^2=162a..........EQ (I) And S2=(27-18)18a=162a........eq (II) S1+S2= distance travelled by the CAR in 27 sec S1+S2=4027=1080.......(iii) Putting the value of S1and S2 from eq (I) and (ii) in eq (iii) 162a+162a=1080 a =1080/324=3.33m/sec^2 Vmax=18*3.33=59.94=60m/sec.

At any instant t = 0 a motorbike start from rest in a given direction, a car moving with constant speed overtakes themotorbike at that instant when it is moving with a speed 40 m/s. Motor bike accelerates uniformly till t = 18s andthen move with constant speed and overtakes the car at t = 27 s.The maximum speed of motorbike in m/s isA) 80(B) 60(C) 100(D) 120plz tell

Answer»

Answer:

60 m/sec

Explanation:

At t=0,u=0for the MOTORBIKE

And at t=18, v=18a(which will be the Vmax)

So, S1=1/2a18^2=162a..........EQ (I)

And S2=(27-18)*18a=162a........eq (II)

S1+S2= distance travelled by the CAR in 27 sec

S1+S2=40*27=1080.......(iii)

Putting the value of S1and S2 from eq (I) and (ii) in eq (iii)

162a+162a=1080

a =1080/324=3.33m/sec^2

Vmax=18*3.33=59.94=60m/sec.

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