sebuah mobil diberi percepatan 2 m/s^2 selama 10 detik dari keadaan diam, kemudian di beri perlambatan sebesar 4 m/s^2 hingga berhenti. jarak total yang ditempu

[tex]\displaystyle a_1=2\,\text{m/s}^2\\v_0=0\,\text{m/s}\\v_t'=0\,\text{m/s}\\t=10\,\text{s}\\a_2=-4\,\text{m/s}\\\\s=?\\\\x_1=v_0t+\frac12a_1t^2\\x_1=0\cdot10+\frac12\cdot2\cdot10^2\\x_1=0+100\\x_1=100\,\text{m}\\\\v_t'^2=v_t^2+2a_2x_2\\v_t'^2=(v_0+a_1t)^2+2a_2x_2\\0^2=(0+2\cdot10)^2+2(-4)x_2\\0=400-8x_2\\8x_2=400\\x_2=50\,\text{m}\\\\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