Questions
A particle starting from rest moves with constant acceleration. If it takes 5 sec to reach the speed 18 km/h. find
a) The average velocity during this period
b) The distance travelled by the particle during this period
detailed solution
Correct option is A
given initial velocity=u=0 final velocity=v=18 km/h=18×518=5m/s time t=5 sec acceleration=a=v−ut=5−05=1m/s2 s=ut+12at2⇒s=12.5m a) Average velocity Vavg=total displacementtime taken=12.55=2.5m/s b) Distance travelled is 12.5mTalk to our academic expert!
Similar Questions
A particle is moving along the x-axis whose acceleration is given by , where x is the location of the particle. At t = 0, the particle is at rest at x = 4/3 m. The distance travelled by the particle in 5 s is
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
