Answer :
The velocity of the lander when t = 5 s is 10 m/s.
To calculate the velocity of the lander when t = 5 s, we first need to determine the impulse applied to the lander by the engine. The impulse-momentum theorem states that the change in momentum of an object is equal to the impulse applied to it. Mathematically, this can be expressed as:
[tex]\[ \Delta p = \int_{0}^{4} T dt \][/tex]
where ([tex]\[ \Delta[/tex] p) is the change in momentum and T is the thrust as a function of time. The mass of the lander is 200 kg.
Given that the thrust varies with time as shown in the graph, we integrate the thrust function over the time interval [0, 4] to find the total impulse:
[tex]\[ \Delta[/tex] p = [tex]\int_{0}^{4} (1600 - 400t) dt[/tex]
[tex]\[ \Delta[/tex] p = [tex]\left[ 1600t - 200t^2 \right]_{0}^{4}[/tex]
[tex]\[ \Delta[/tex] p = 6400 - 3200 = 3200Ns
Since impulse is equal to the change in momentum, we have:
[tex]\[ \Delta[/tex] p = m [tex]\[ \Delta[/tex] v
3200 = 200 [tex]\[ \Delta[/tex] v
[tex]\[ \Delta[/tex] v = 3200 / 200 = 16 m/s
The initial velocity of the lander is 6 m/s downward, so the final velocity when t = 5 s is:
v[tex]_f[/tex] = 6 - 16 = -10 m/s
Therefore, the velocity of the lander when t = 5 s is 10 m/s.
The complete question would be:
The 200-kg lunar lander is descending onto the moon's surface with a ve/ocity of 6 m/s when its retro-engine is fired. If the engine produces a thrust T for 4 seconds which varies with time as shown and then cuts off (i.e., T = 0 for t > 4 (s)), calculate the velocity of the lander when t = 5 (s), assuming that it has not yet landed. Gravitational acceleration at the moon's surface is 1.62m/s2.
