a) A motorboat traveling on a straight course slows uniformly from 70 km/h to 40

Question

a) A motorboat traveling on a straight course slows uniformly from 70 km/h to 40 km/h in a distance of 55 m .

1a) What is the magnitude of the boat’s acceleration?

b) A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.50 m/s2 . At 10.0 s after blastoff, the engines suddenly fail, which means that the force they produce instantly stops.

1b) How long after it was launched will the rocket fall back to the launch pad?

2b) How fast will it be moving when it does so?

Explanation / Answer

a) The acceleration is constant here so we can use the third equation of motion to find out the acceleration.

v^2 = u^2 + 2as

v = 40 km/h = 11.11 m/s

u = 70 km/h = 19.44 m/s

a = (11.11^2-19.44^2)/(2*55) = -2.31 m/s^2

magnitude of acceleration a = 2.31 m/s^2

b) 1. How long after it was launched will the rocket fall back to the launch pad?

time taken before engine stops t1 = 10 sec

After 10s it has velocity v = at = 2.5*10 = 25 m/s

hence time taken to reach maximum height t2 = 25/9.81 = 2.55 sec.

the maximum height = 0.5at1^2 + v^2/(2g) = 0.5*2.5*10^2 + 25^2/(2*9.81) = 156.86 m

time taken to reach the pad again t3 = sqrt(2h_max/g) = sqrt(2*156.86/9.81) = 5.66sec

total time = t1+t2+t3 = 18.21 seconds

2. The velocity at the pad = g*t3 = 9.81*5.66 = 55.52 m/s

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