Sebuag mobil diberi percepatan 2 m/s² selama 10 detik dari keadaan diam, kemudian diberi perlambatan sebesar 4m/s² hingga berhenti. Jarak total yg ditempuh mobi

[tex]\displaystyle a_1=2\,\text{m/s}^2\\t=10\,\text{s}\\v_0=0\,\text{m/s}\\a_2=-4\,\text{m/s}^2\\v_t'=0\,\text{m/s}\\\\s=?\\\\v_t=v_0+a_1t\\v_t=0+2\cdot10\\v_t=20\,\text{m/s}\\\\v_t^2=v_0^2+2a_1x_1\\20^2=0^2+2\cdot2x_1\\400=4x_1\\100\,\text{m}=x_1\\\\v_t'^2=v_t^2+2a_2x_2\\0'^2=20^2+2\cdot(-4)x_2\\0'^2=400-8x_2\\50\,\text{m}=x_2\\\\s=x_1+x_2\\s=100+50\\\boxed{\boxed{s=150\,\text{m}}}[/tex]

Gerak Lurus.
Kelas X kurikulum 2013 revisi 2016.

Δx₁ = v₀t₁ + 1/2 a₁t₁²
       = 0 + 1/2 (2 m/s²)(10 s)² = 100 m
v₁ = v₀ + a₁t₁
    = 0 + (2 m/s²)(10 s) = 20 m/s

v₂² = v₁² + 2a₂Δx₂
Δx₂ = (v₂² - v₁²) / (2a₂)
       = [0 - (20 m/s)²] / [2(-4 m/s²)] = 50 m

Δx = Δx₁ + Δx₂
     = 100 m + 50 m = 150 m