Question

A body starts from rest and moves with uniform acceleration of 2 m/s2 for 10 sec, it move with constant speed for 30 sec then decelerates by 4 m/s2 to zero. What is the total distance covered by the body? -​

Answer 1

u = 0

a = 2m/S²

t = 10 Seconds

S₁​ = ut + 0.5at²

s₁​ = 0 + 0.5 × 2 × 100 = 100m

S₂ = v + 2

v = u + at = 0 + 20 = 20m/s

s₂ = 20 × 30 = 600m

0² − v² = 2( −a)s₃

s₃ = 400/8 = 50m

s = s₁ + s₂ + s₃ = 100 + 600 + 50 = 750m

Answer 2

Answer:

$distance \: travelled \: by \: the \: body \: during \\ uniform \: acceleration \\ s = ut + \frac{1}{2}a {t}^{2} \\ s = \frac{1}{2} \times 2 \times {10}^{2} \\ s = 100m \\ final \: velocity \: of \: the \: body for \: 10sec \\ v = u + at \\ v = 0 + 2 \times 10 \\ v = 20 \frac{m}{s} \\ distance \: covered \: by \: the \: body \: when \\ the \: body \: moves \: with \: constant \: velocity \\ s = ut + \frac{1}{2}a {t}^{2} \\ s = 20 \times 30 + 0 \\ s = 600m \\ time \: taken \: by \: the \: body \: to \: decelerate \: \\ and \: come \: to \: rest \\ v = u + at \\ 0 = 20 - 4t \\ t = 5sec \\ distance \: travelled \: by \: the \: body \: during \\ deceleration \\ s = ut + \frac{1}{2}a {t}^{2} \\ s = 20 \times 5 - \frac{1}{2} \times 4 \times {5}^{2} \\ s = 50m$

Total distance travelled : 100 + 600 + 50 → 750m

Hope it's helpful to all