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1) a car travelling at 50km per second from rest cover a distance of 10km in 40 minu,caculate the acceleration of car.

2) a car accelerate uniformly from rest at 5 meter per second square.determine it speed after 10 second.
3) a body is dropped from a height above a ground for a velocity after 4 second take g as 9.8mls square 4) a bird fly at 10mls for 3 second,15mls for 3 second and 20mls for 14 second .caculate the average speed of the bird.
5) a bell is throne vertically upward an initial velocity of 40 mls find (a) its velocity after 3 second (b) the maximum height attend and the time taken to reach
6)a body accelerate uniform rest at 2mls square.calculate the magnitude of it velocity after travelling in 9 meter

Post your attempts.

accelerate for time = t
cruise at 50 m/s (surely not at 50 km/s unless car is in orbit, check your typing for typos. Maybe 50 km / hour which is about 30 miles/hour ?)
cruise at 50 for time T
t + T = 40 min = 2,400 seconds
steady speed = 50 whatevers/whatever = 0 + a t
distance during acceleration = d = .5 a t^2
total distance = 40 dowhistles = d + D where D is distance at steady speed
so D = 50 T
40 = .5 a ^2 + 50 T

2) v = 0 + a t = 500

forgot decimal point

50.0

Did you know?

Did you know that acceleration is a measure of how quickly an object's velocity changes? In the first scenario, if a car is initially at rest and travels 10km in 40 minutes, we can calculate its acceleration. By converting 40 minutes to 40/60 = 2/3 hours and dividing 10km by 2/3 hours, we get a velocity of 15km/h. As the car traveled from rest to 15km/h in 2/3 hours, the acceleration can be calculated as (15km/h) / (2/3 hour) = 22.5 km/h^2. In the second scenario, if a car accelerates uniformly from rest at 5m/s^2, we can determine its speed after 10 seconds. Using the equation v = u + at, where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time, we can calculate the speed as v = 0 + (5m/s^2) * (10s) = 50m/s. In the third scenario, a body is dropped from a height above the ground. After 4 seconds, we can determine the velocity using the acceleration due to gravity, which is approximately 9.8m/s^2. Using the equation s = ut + (1/2)at^2, where s is the distance (height) traveled, u is the initial velocity (which is 0 in this case), a is the acceleration due to gravity, and t is the time, we can calculate the velocity as s = (1/2) * (9.8m/s^2) * (4s)^2 = 78.4m/s. In the fourth scenario, to calculate the average speed of a bird that flies at different speeds for different durations, we can add up the distances traveled and divide it by the total time taken. The bird flies at 10m/s for 3 seconds, covering a distance of 10m. It then flies at 15m/s for 3 seconds, covering a distance of 45m. Lastly, it flies at 20m/s for 14 seconds, covering a distance of 280m. Adding up the distances (10m + 45m + 280m) and dividing it by the total time (3s + 3s + 14s), we find that the average speed of the bird is (335m) / (20s) = 16.75m/s. In the fifth scenario, when a bell is thrown vertically upward with an initial velocity of 40m/s, we can determine its velocity after 3 seconds using the equation v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time. So, v = 40m/s + (-9.8m/s^2) * (3s) = 10.6m/s. The maximum height attained by the bell can be calculated using the equation s = ut + (1/2)gt^2, which simplifies to s = (1/2) * (-9.8m/s^2) * (3s)^2 = 44.1m. To find the time taken to reach the maximum height, we use the equation v = u + gt, setting v = 0 to find when the upward velocity is reduced to zero. Therefore, 0 = 40m/s + (-9.8m/s^2) * t. Solving for t gives t = 4.081s. In the sixth scenario, if a body accelerates uniformly from rest at 2m/s^2, we can calculate the magnitude of its velocity after traveling 9m using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration, and s is the distance traveled. Plugging in the values, we have v^2 = 0^2 + 2 * (2m/s^2) * (9m) = 36m^2/s^2. Taking the square root of both sides, we find that the magnitude of the velocity is v = 6m/s.