Question

 A car going 5 m/s accelerates at 2 m/s^2 for 3 seconds.  Then he travels at a constant velocity for 4 seconds. What is the total distance traveled in the 7 seconds and what is the final speed of the car?

Answer 1

Answer 2

We can break down the motion of the car into two parts: the first 3 seconds where it is accelerating and the next 4 seconds where it is moving at a constant velocity.

For the first part, we can use the kinematic equation: d = vit + 1/2at^2 where d is the distance traveled, vi is the initial velocity, t is the time, and a is the acceleration. Substituting the given values, we get: d = (5 m/s)(3 s) + 1/2*(2 m/s^2)*(3 s)^2 = 22.5 m

For the second part, since the car is moving at a constant velocity, we can use the formula: d = vt where v is the constant velocity and t is the time. Substituting the given values, we get: d = (5 m/s + 2 m/s^23 s)*(4 s) = 28 m

Therefore, the total distance traveled in the 7 seconds is: d_total = 22.5 m + 28 m = 50.5 m

To find the final speed of the car, we can use the formula: v_f = v_i + at where v_i is the initial velocity, a is the acceleration, and t is the time. Substituting the given values, we get: v_f = 5 m/s + 2 m/s^23 s = 11 m/s

Therefore, the final speed of the car is 11 m/s.