Respuesta :
Answer:
in first case velocity =21.07 m/sec in second case velocity =15.05 m/sec
Explanation:
We have given initial velocity u = 7.9 m/sec
Acceleration [tex]a=0.72m/sec^2[/tex]
Distance S = 265 m
Now according to third law of motion [tex]v^2=u^2+2as[/tex] here v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance
So [tex]v^2=7.9^2+2\times 0.72\times 265=444.01[/tex]
v = 21.07 m/sec
In second case s =114 m
So [tex]v^2=7.9^2+2\times 0.72\times 114=226.57[/tex]
v =15.05 m/sec
Answer:
1. 21.07 m/s
2. 15.05 m/s
Given:
initial speed of the car, u = 7.9 m/s
distance covered by the car, d = 265 m
acceleration, a = 0.72[tex]m/s^{2}[/tex]
Solution:
To calculate the velocity at the end of acceleration, we use the third eqn of motion:
[tex]v^{2} = u^{2} + 2ad[/tex]
[tex]v^{2} = 7.9^{2} + 2\times 0.72\times 265[/tex]
[tex]v = \sqrt{7.9^{2} + 2\times 0.72\times 265} = 21.07 m/s[/tex]
Now,
Velocity after it accelerates for a distance for 114 m:
Here d = 114 m
Again, from third eqn of motion:
[tex]v^{2} = u^{2} + 2ad[/tex]
[tex]v = \sqrt{7.9^{2} + 2\times 0.72\times 114} = 15.05 m/s[/tex]
