6- The acceleration of an object as a function of time is given by oft-(2.00 ms

Question

6- The acceleration of an object as a function of time is given by oft-(2.00 ms If the objoct is at rest at time 0.00s, what is the velocity of the objost at time t 5.o00 s7 A) 25.0ms ) 37.5ms C) 0.00 ms D) 12.0 m's E) 75.0 ms 7- Two objects are dropped from a bridgc, an interval of 2.0 s apart, and experience no appreciable air resistance. As time progressos, the DIPPERENCE In their speeds A) Increases at first, but then stays constant B) Remains constant C) Decreases at first, but then stays constant D) Increases E) Decreases Object A has a position as a function of time givon by FA-(3.00 ms+ (1.00ms Object B has a position as a functon oftme given by rain-(4.00 m/s)(-100 nasyy All quantities are SI units. What is the distance betwoem object A and objoct B at time 3.84 s? A) 34.6 m B) 15.0m C) 18.3 m D) 3.46m E) 29.7m 9. A boy throws a rock with an initial velocity of 5.73 mis at 30.0 above the horizontal. If air resistance is negligible, how long does it take the rock to reach the maximum height of its A) 0.110 s B) 0.194 s C) 0.215 s D) 0.292 E)0.303

Explanation / Answer

6) Integrate the acceleration wrt time to get velocity. V = initial velocity + 2*(t^2)/2 = 0 + t^2 = t^2

Velocity at t = 5 is equal to 25 m/sec

7) rate of change of relative speed is given by difference in their acceleration.

d(V1 - V2)/dt = dv1/dt - dv2/dt = aceeleration of 1 - acceleration of 2 = g -g = 0

So the difference remains constant.

8) rA - rB = -1t + 2t^2

at t = 3.84 sec, rA-rB = -3.84i + 29.5j

Distance = sqrt (3.84^2 + 29.5^2) = 29.7 m

9) Vertical velocity is 5.73 * sin(30) m/s

Acceleration is -9.8 m/s2

time = (0-velocity)/acceleration

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