Q3 A car A is travelling along a straight path, while a lorry B is travelling along a circular path having a radius of curvature of 200 m as shown in Figure Q3. At the instant shown, car A travel at the velocity of (150+X) km/h while lorry B travel at the velocity of 80 km/h. Also at this instant, car A has an increase velocity of 7000 km/h² and lorry B has a decrease velocity of 4500 km/h. The angle between straight path of car A and the horizontal line is 0=65°. (а) Calculate the magnitude and direction of velocity of lorry B with respect to car A. Determine the magnitude and direction of acceleration of lorry B as measured by the (b) driver of car A. Use your matrix number by adding the last three digits to get the value of X on the velocity of car A. The value of X will then need to be added with the digit of 150 as follows; Example; If your matrix number is AD190251, Velocity of car A, VA = 150 + X km/h X= (2 + 5 + 1) = 8 Therefore, the velocity of car A, VA = 150 + X, = 150 + 8 = 158 km/h.

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Q3 A car A is travelling along a straight path, while a lorry B is travelling along a circular path
having a radius of curvature of 200 m as shown in Figure Q3. At the instant shown, car A
travel at the velocity of (150+X) km/h while lorry B travel at the velocity of 80 km/h. Also at
this instant, car A has an increase velocity of 7000 km/h² and lorry B has a decrease velocity
of 4500 km/h. The angle between straight path of car A and the horizontal line is 0=65°.
(a)
Calculate the magnitude and direction of velocity of lorry B with respect to car A.
(b)
Determine the magnitude and direction of acceleration of lorry B as measured by the
driver of car A.
Use your matrix number by adding the last three digits to get the value of X on the velocity
of car A. The value of X will then need to be added with the digit of 150 as follows;
Example; If your matrix number is AD190251,
Velocity of car A, VA=150 + X km/h
X= (2 + 5 + 1) = 8
Therefore, the velocity of car A, VA = 150 + X, = 150 + 8 = 158 km/h.
Transcribed Image Text:Q3 A car A is travelling along a straight path, while a lorry B is travelling along a circular path having a radius of curvature of 200 m as shown in Figure Q3. At the instant shown, car A travel at the velocity of (150+X) km/h while lorry B travel at the velocity of 80 km/h. Also at this instant, car A has an increase velocity of 7000 km/h² and lorry B has a decrease velocity of 4500 km/h. The angle between straight path of car A and the horizontal line is 0=65°. (a) Calculate the magnitude and direction of velocity of lorry B with respect to car A. (b) Determine the magnitude and direction of acceleration of lorry B as measured by the driver of car A. Use your matrix number by adding the last three digits to get the value of X on the velocity of car A. The value of X will then need to be added with the digit of 150 as follows; Example; If your matrix number is AD190251, Velocity of car A, VA=150 + X km/h X= (2 + 5 + 1) = 8 Therefore, the velocity of car A, VA = 150 + X, = 150 + 8 = 158 km/h.
v- (150 + X) km/h
Vg- 80 km/h
as= 7000 km/h?
B
ag- 4500 km/h² \
Figure Q3
Transcribed Image Text:v- (150 + X) km/h Vg- 80 km/h as= 7000 km/h? B ag- 4500 km/h² \ Figure Q3
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