Question

The graph shows the speed of a car as a function of time 16 13 12 10 0123456789 10 t (s) Initially the car is at rest. What is the acceleration of the car? Please, note that the graph goes through at least one grid intersection point. 1.556 m/s2 You are correct. our receipt no. is 155-1116 Previous Tries How much distance does the car cover between t1 2.19 s and t2 5.55 s? 25.958 m

Answer 1

The velocity-time graph of an object represents the speed at which an object is moving at a given time. Objects move at variable speeds have sloping, linear velocity graphs. The slope of the velocity-time graph represents the acceleration of an object. When the slope of the velocity-time graph is a horizontal line, the acceleration will be zero. This means the object is either at rest or moving at a constant speed, without speeding up or slowing down Refer to the given graph, the slope intersect at a point (10,15.56). This coordinates represents the velocity of the car is 15.56 m/s in time 10 s. Find the acceleration using the kinematic equation of motion. v=u+at Here, v is the final velocity, u is the initial speed, a is the acceleration and t is the time. Since the initial speed of the car is zero. Thus, the above equation becomes, v=at Substitute 15.56 m/s for v and 10 s fort in the above equation. 15.56 m/s = a(10.0 s) Solve the above equation for the acceleration of the car. a=1.556 m/s Hence, the acceleration of the car is 1.556 m/s?.

The velocity of the car is given by, Av = 8.0 m/s-4.0 m/s = 4.0 m/s The time taken by the car is given by, At = 5.55 s-2.19 s = 3.36 s The total distance cover by the car is given by, d=avat +-aar? = (4.0 m/s)(3.36 s) + (1.556 m/s^)(3.365)* = 13.44 m+8.78 m = 22.22 m Hence, the total distance cover by the car is (22.22 m