A helicopter is rising straight up in the air. Its distance from the ground t se

Question

A helicopter is rising straight up in the air. Its distance from the ground t seconds after takeoffis s() feet, where s()-2? +6t (a) How long will it take for the helicopter to rise to 20 feet? (b) Find the velocity and acceleration of the helicopter when it is 20 feet above the ground. (a) The time it will take for the helicopter to rise to 20 feet isseconds. (b) When the helicopter is 20 feet above the ground, its velocity is feet per second. When the helicopter is 20 feet above the ground, its acceleration is feet per second squared. Enter your answer in each of the answer boxes

Explanation / Answer

The distance from the ground 't' seconds after take-off is s(t) = 2t2 + 6t;
So for a height of 20 feet, s(t) = 20;
2t2 + 6t = 20
2t2 + 6t - 20 =0;
2t2 + 10 t - 4t - 20=0
2t ( t+5) - 4 (t+5) =0
(2t-4) ( t+5) =0
so 2t-4=0 meaning 2t=4 and t=4/2 = 2;
or t+ 5 =0 or t=-5 ;
t=-5 is not possible since time cannot be negative, so t= 2 seconds;
Thus, the time helicopter will take to rise to 20 feet is = 2 seconds

Velocity is given by rate of change of distance or v(t) = d s(t)/dt
v(t) = d/ dt ( 2t2 + 6t) = 4t + 6;
at s(t) = 20 feet we got t=2 seconds, hence velocity at t=2 seconds is v(2) = 4(2) + 6 = 8+6 = 14 feet/sec


Acceleration is given by rate of change of velocity, a(t) = d / dt ( v(t) )
a(t) = d/dt ( 4t+6) = 4
At s(t) = 20 feet, we got t=2 seconds, so acceleration at t=2 seconds is a(t) = 4 fee/ sec squared

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