The velocity-:me graph of a particle moving along a straight line is given by graph shown The rate of acceleration and deceleration is same and it is equal to 4 m/s2. If the average velocity during the motion is 15 m/s and total :me of motion is 20 seconds then find (a) the value of t (b) the maximum velocity of the particle during the journey. (c) the distance travelled with uniform velocity.

The average velocity (vₐ) is the total displacement (s) divided by the total time (T). On a velocity-time graph, the total displacement is equal to the area under the curve. In this case, the region under the curve is a trapezoid. The area of a trapezoid is the average of the parallel sides (a and b) times the perpendicular height (h). So the total displacement of the particle is:

s = ½ (a + b) h

s = ½ (20 − 2t + 20) v

s = (20 − t) v

The acceleration is the slope of the line. Given that the acceleration is 4 m/s², the maximum velocity is v = 4t. Substituting, the displacement is:

s = (20 − t) 4t

s = 80t − 4t²

Given that the average velocity is 15 m/s:

vₐ = s / T

15 = (80t − 4t²) / 20

300 = 80t − 4t²

4t² − 80t + 300 = 0

t² − 20t + 75 = 0

(t − 5) (t − 15) = 0

t = 5 or 15

Since 20 − 2t must be positive, t can't be 15. Therefore, t = 5 seconds.

Part b

The maximum velocity is:

v = 4t

v = 20 m/s

Part c

The distance traveled with uniform velocity is:

d = v (20 − 2t)

d = 200 m