a train started from rest and moving with constant acceleration upto 10 sec. after 5sec its speed become 42 km/h .then
(1) find acceleration of train,,
(2)speed after 10 sec
(3)distance travelled by train in 7 sec and10 sec respectively
Answers
Given :
- initial velocity of train , u = 0
- time for which train moved with constant acceleration , t = 10 s
- velocity of train after 5 sec (t') , v = 42 km/h
velocity of train after 5 sec , v = 35/3 m/s
To find :
- Acceleration of Train , a = ?
- speed of train after 10 sec , V = ?
- distance travelled by train in t₁ = 7 sec and t₂ = 10 sec respectively
Formula required :
- First equation of motion
v = u + a t
(where v is the final velocity , u is the initial velocity , a is the acceleration and t is the time taken)
- second equation of motion
s = u t + 1/2 a t²
(where u is the initial velocity , s is the distance travelled , a is acceleration and t is time taken)
Solution :
Using first equation of motion
→ v = u + a t'
→ 35/3 = 0 + a (5)
→ a = 7/3 m/s²
therefore,
Acceleration of Train is 7/3 m/s² .
Again ,
using first equation of motion
→ V = u + a t
→ V = 0 + 7/3 × 10
→ V = 70/3 m/s
therefore,
Speed of train after 10 sec is 70/3 m/s .
Now,
Using second equation of motion
→ s₁ = u t₁ + 1/2 a t₁²
→ s₁ = (0)(7) + 1/2 × (7/3) × (7)²
→ s₁ = 343/6 m
Therefore,
Distance travelled by train in 7 sec is 343/6 m .
Again,
Using second equation of motion
→ s₂ = u t₂ + 1/2 a t₂²
→ s₂ = (0)(10) + 1/2 × (7/3) × (10)²
→ s₂ = 350/3 m
Therefore,
Distance travelled by train in 10 sec is 350/3 m .