A body travels 200cm in the first two seconds and 220cm in next four s...
For first 2 seconds,S=ut+1/2at^2.200=2u+1/2 a 4.200=2u+2a.100=u+a.-----(1)for next 4 seconds,the distance travelled becomes=200+220=420cm.and time becomes=2+4=6sec.S=ut+1/2at^2.420=6u+1/2 a 36.420=6u+18a.70=u+3a.------(2).on solving( 1 ) and (2) equation.we get,u=115cm/s,a=-15cm/s^2.V=u+at.V=115+(-15)(7).V=115-105.V=10cm/s....
A body travels 200cm in the first two seconds and 220cm in next four s...
Given Information:
- The body travels 200cm in the first two seconds.
- The body travels 220cm in the next four seconds with deceleration.
Calculating the Acceleration:
Let's assume the initial velocity of the body is 'u' cm/s and the acceleration is 'a' cm/s².
Using the first equation of motion, we have:
Distance = Initial Velocity × Time + (1/2) × Acceleration × Time²
For the first two seconds:
200 = u × 2 + (1/2) × a × (2)²
200 = 2u + 2a ---(1)
For the next four seconds:
220 = u × 4 + (1/2) × a × (4)²
220 = 4u + 8a ---(2)
Solving equations (1) and (2), we get:
2u + 2a = 200 ---(3)
4u + 8a = 220 ---(4)
Multiplying equation (3) by 2 and subtracting it from equation (4), we get:
4u + 8a - (4u + 4a) = 220 - 400
4a = -180
a = -45 cm/s²
Calculating the Final Velocity:
We can use the third equation of motion to calculate the final velocity.
Final Velocity = Initial Velocity + (Acceleration × Time)
For the seventh second:
Final Velocity = u + (a × 7)
Final Velocity = u - 45 × 7
Final Velocity = u - 315
Since the body is decelerating, the acceleration is negative. Therefore, the final velocity at the end of the 7th second is u - 315 cm/s.
Answer:
The final velocity of the body at the end of the 7th second is u - 315 cm/s.
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