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

A plane has a take-off speed of 120 ms1 after covering a distance of 2400 m. Determine the acceleration (a) of the plane (assuming it to be constant) and the time (t) required to reach that speed.

A
a=4ms2,t=30 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
a=3ms2,t=30 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
a=4ms2,t=40 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
a=3ms2,t=40 s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D a=3ms2,t=40 s
Given:
Initial velocity, u=0 ms1
Final velocity, v=120 ms1
Distance covered, s=2400 m

Let 'a' be the acceleration
Let 't' be the time taken

By the third equation of motion:
v2=u2+2as
1202=02+2×a×2400
14400=4800a
a=3 ms2

By the first equation of motion:
v=u+at
120=0+3t
t=40 s

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Equations of Motion - concept
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon