CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Two particles, one with constant velocity 50 ms1 and the other with uniform acceleration 10 ms2, start moving simultaneously from the same place in the same direction. When will the accelerated particle be 125 m ahead of the other particle?

A
3 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
5(1+2) s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
10 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
10(2+1) s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 5(1+2) s
For particle 1 moving with constant velocity, S1=50t ... (i)
For particle 2 moving with constant acceleration,
S2=12at2=12×10t2=5t2 ... (ii)
As per question;
S2S1=125, we get
From equations (i) & (ii)
5t250t=125
t210t25=0
On solving, we get
t=(5±52) s
t=5(1+2) s (taking positive value)

Alternate Solution:
Relative displacement of particles 1 and 2 is
S12=125 m
Initial velocity of 1 w.r.t. 2,
U12=500=50 m
Acceleration of 1 w.r.t. 2
a12=010=10 m/s2
Using second equation of motion,
S12=U12t+12a12t2
125=50t+12×(10)t2
t210t25=0
t=5(1+2) or t=5(12) sec
Taking +ve value
t=5(1+2) s

flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon