Question

Refer to the attachment. ​

Answer 1

A = (v - u)/t

= (0 - u)/t

= -u / (5 s)

S = ut + 0.5at²

100 = 5u + (0.5 × [-u / (5 s)] × (5 s)²)

100 = 5u - 2.5u

100 = 2.5u

u = 40 m/s

If u = 40 m/s then

a = -u / (5 s)

= -(40 m/s) / (5 s)

= -8 m/s²

Displacement covered in 10 seconds

S = ut + 0.5at²

= (40 m/s × 10 s) + [0.5 × (-8 m/s² × (10 s)²]

= 0 m

Particle travelled 100 m in first 5 seconds and in next 5 seconds it travels 100 m back to initial position. Hence we got Displacement after 10 seconds equal to zero.

Answer 2

Answer: 140 m/s

Explanation: Change in velocity= Area under acceleration-time graph=

Area of trapezium = $\frac{1}{2} *(10+4) * 20$ = 140 m/s