Question

question 2 (1 point)
a car starting from rest moves with constant acceleration of 2.0 m/s2 for 10 s, then travels with constant speed for another 10 s, and then finally slows to a stop with constant acceleration of -2.0 m/s2. how far does it travel?
200 m
300 m
400 m
500 m

Explanation:

Step1: Find velocity after acceleration

Use \( v = u + at \). Initial velocity \( u = 0 \), \( a = 2.0 \, \text{m/s}^2 \), \( t = 10 \, \text{s} \).
\( v = 0 + 2.0 \times 10 = 20 \, \text{m/s} \).

Step2: Distance during acceleration (\( d_1 \))

Use \( d = ut + \frac{1}{2}at^2 \). \( u = 0 \), so \( d_1 = 0 + \frac{1}{2} \times 2.0 \times 10^2 = 100 \, \text{m} \).

Step3: Distance during constant speed (\( d_2 \))

Speed \( v = 20 \, \text{m/s} \), time \( t = 10 \, \text{s} \).
\( d_2 = v \times t = 20 \times 10 = 200 \, \text{m} \).

Step4: Distance during deceleration (\( d_3 \))

Final velocity \( v_f = 0 \), initial velocity \( v = 20 \, \text{m/s} \), \( a = -2.0 \, \text{m/s}^2 \).
Use \( v_f^2 = v^2 + 2ad_3 \). Solve for \( d_3 \):
\( 0 = 20^2 + 2(-2.0)d_3 \)
\( 4d_3 = 400 \)
\( d_3 = 100 \, \text{m} \).

Step5: Total distance (\( d_{\text{total}} \))

\( d_{\text{total}} = d_1 + d_2 + d_3 = 100 + 200 + 100 = 400 \, \text{m} \).

Answer:

400 m