An object is moving with uniform acceleration.Its velocity after 5s is 25m/s and after 8s is 34m/s.Find the distance travelled by the object in 12th second.
Answers
Answered by
255
v = u + at
After 5s
25 = u + 5a ——(1)
After 8s
34 = u + 8a ——(2)
Subtract equation (1) from equation (2)
34 - 25 = u + 8a - u - 5a
9 = 3a
a = 3 m/s^2
Substitute a = 3m/s^2 in equation (1)
25 = u + 5*3
u = 10 m/s
Distance travelled in 12th second
S = u + a(n - 1/2)
S = 10 + 3*(12 - 1/2)
S = 44.5m
Distance travelled in 12th second is 44.5m
After 5s
25 = u + 5a ——(1)
After 8s
34 = u + 8a ——(2)
Subtract equation (1) from equation (2)
34 - 25 = u + 8a - u - 5a
9 = 3a
a = 3 m/s^2
Substitute a = 3m/s^2 in equation (1)
25 = u + 5*3
u = 10 m/s
Distance travelled in 12th second
S = u + a(n - 1/2)
S = 10 + 3*(12 - 1/2)
S = 44.5m
Distance travelled in 12th second is 44.5m
Answered by
40
First we need calculate the acceleration so we find the difference between the two velocity equations.34-25=9 and 8-5=3.so it is 9 m/s in 3 second,now we can calculate the acceleration.
A = V - U/T => 9/3 = 3 m/s^2
so starting from the 5th second after 7 seconds the velocity = 25 + 7 × 3 = 46m/s
So the velocity in the twelfth second is 46m/s.
Distance = Average Velocity × Time
=>46/2 × 12 = 276 metres.
A = V - U/T => 9/3 = 3 m/s^2
so starting from the 5th second after 7 seconds the velocity = 25 + 7 × 3 = 46m/s
So the velocity in the twelfth second is 46m/s.
Distance = Average Velocity × Time
=>46/2 × 12 = 276 metres.
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