1.Moving with uniform acceleration a body covers 150m during 10sec so thet it covers 24m during the 10th sec .What is the initial velocity and the acceleration of the body.
2.A runner travels 1.5 laps around a circular track of 50sec .The diameter of the track is 40m and its circumferenceis 126m.Find a)average speed of the runner and b)the magnitude of the runner average velociy
2 answers
Someone will gladly critique your thinking, when any is shown.
1.
s =vₒ•t +a•t²/2,
v(9) = vₒ +a•9,
24 =v(9) •1 + a•1²/2 =
=vₒ +a•9 + a•1²/2 =
= vₒ +9.5•a.
s =vₒ•t +a•t²/2 =>
150 = vₒ•10+ a•100/2 =
= 10• vₒ + 50•a.
24= vₒ +9.5•a,
150= 10• vₒ + 50•a.
45° = 90,
a = 2 m/s².
vₒ = 24-9.5 •a = 5 m/s.
2.
Average speed = total distance travelled/ total time taken = 1.5•126/50 = 3.78 m/s,
Displacement is the distance from the starting point to the end point.
Average velocity = final displacement/total time taken = D/t = =40/50 =0.8 m/s.
s =vₒ•t +a•t²/2,
v(9) = vₒ +a•9,
24 =v(9) •1 + a•1²/2 =
=vₒ +a•9 + a•1²/2 =
= vₒ +9.5•a.
s =vₒ•t +a•t²/2 =>
150 = vₒ•10+ a•100/2 =
= 10• vₒ + 50•a.
24= vₒ +9.5•a,
150= 10• vₒ + 50•a.
45° = 90,
a = 2 m/s².
vₒ = 24-9.5 •a = 5 m/s.
2.
Average speed = total distance travelled/ total time taken = 1.5•126/50 = 3.78 m/s,
Displacement is the distance from the starting point to the end point.
Average velocity = final displacement/total time taken = D/t = =40/50 =0.8 m/s.
1 answer