Davien and Jackson are headed to a beach that is 50km from school. At noon, Davien leaves his house that is 10 km closer to the
1 answer:
Distance between Davien's school and beach, D = 50 km.
Also, Davian's house is 10 km closer to the beach than the school.
So, distance between Davien's house and beach, d = 40 km.
Speed of Davian, s = 40 km/h.
Now, time taken is given by :
Therefore, Davien reach the beach at 1 : 30 P.M .
Hence, this is the required solution.
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<u>Answer:</u>
a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s
b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s
<u>Explanation:</u>
a) The height of ramp = 1.5 meter
Horizontal distance he must clear = 22 meter
The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.
We have equation of motion, , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 , we need to calculate time when displacement = 1.5 meter.
So the car has to cover a distance of 22 meter in 2.119 seconds.
So minimum speed required = 22/0.553 = 39.78 m/s
Minimum speed must he drive off the horizontal ramp = 39.78 m/s
b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V
So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8
Substituting
In case of horizontal motion
Horizontal speed of car = V cos 7 = 0.993V
So it has to travel 22 meter in t seconds
0.993Vt = 22, Vt = 22.155 m
Substituting in the equation
We will get
Speed required = 22.155/0.926 = 23.93 m/s
Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s