At time t1 = 16 s, a car with mass 1000 kg is located at <112,...
At time t1 = 16 s, a car with mass 1000 kg is located at <112, 0, 20> m and has momentum <5500, 0, −3600> kg · m/s. The car's momentum is not changing. At time t2 = 20 s, what is the position of the car?
Homework Answers
Solution:
The car's momentum is not changing.
This implies that the dp/dt = 0
=> Force and hence acceleration = 0
Velocity of the car at the time t1 = 16 s is momentum / mass
v = < 5500 , 0 , -3600 > /(1000) = < 5.5 , 0 , -3.6 >
Displacement of the car , when velocity is constant = X = vt
= <5.5 ,0 , -3.6 > * ( 20 - 16 ) = < 5.5 *4 , 0 , -3.6 * 4>
= < 112 + 22, 0 , 20- 14.4 > m
=> Position of the car at t=20 s = < 134 , 0 , 5.6 > m
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Table 3.1 Time, t (s) Position, x (m) tn/t1 Time Squared, t^2 (s^2) (tn)^2/(t1)^2 xn/x1 t1,...
Table 3.1
Time, t (s)
Position, x (m)
tn/t1
Time Squared, t^2 (s^2)
(tn)^2/(t1)^2
xn/x1
t1, 0.40
x1, 0.25
1.0
.16
1
1
t2, 0.80
x2, 0.36
2.0
.64
4
1.4
t3, 1.2
x3, 0.52
3.0
1.4
9
2.1
t4, 1.6
x4, 0.71
4.0
2.6
16
2.8
t5, 2.0
x5, 0.95
5.0
4.0
25
3.8
t6, 2.4
x6, 1.2
6.0
5.8
36
4.8
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Table 3.1
Time, t (s)
Position, x (m)
tn/t1
Time Squared, t^2 (s^2)
(tn)^2/(t1)^2
xn/x1
t1, 0.40
x1, 0.25
1.0
.16
1
1
t2, 0.80
x2, 0.36
2.0
.64
4
1.4
t3, 1.2
x3, 0.52
3.0
1.4
9
2.1
t4, 1.6
x4, 0.71
4.0
2.6
16
2.8
t5, 2.0
x5, 0.95
5.0
4.0
25
3.8
t6, 2.4
x6, 1.2
6.0
5.8
36
4.8
Linear fit equation for Position Vs. Time: 0.545x - 0.0530
Quadratic fit equation for Position Vs. Time:...
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