Physics Jamb, Waec, Neco Past Questions & Answers on Speed, Velocity and Acceleration

QUESTIONS 21. A particle moving in a straight line with uniform deceleration has a velocity of 40m/s at a point P, 20m/s at a point Q and comes to rest at a point R where QR=50m. 

Calculate the: 

(i) distance PQ; 

(ii) time taken to cover PQ; 

(iii) time taken to cover PR. 

WAEC1990 Ans: (i) 150m; (ii) 5s (iii) 10s 

Solutions 

QR = s = 50m

VQ = u = 20m/s

VR = v = 0m/s

Using 

v² = u² + 2as for acceleration a

0² = 20² + 2 (a) (50)

0 = 400 + 100a

100a = -400

a = -400/100

a = - 4m/s²

I) PQ = s = ?

Solution 

Vp = u = 40m/s

Vq = v = 20m/a 

a = -4m/s²

Using 

V² = u² + 2as

20² = 40² + 2 (-4)s

400 = 1600 -8s

8s = 1600 -400

8s = 1200

s = 1200/8

s =150m ; therefore PQ = 150m

Il) Time taken t to cover PQ 

v= 20m/s

u = 40m/s

a = -4m/s²

t = ?

Using v = u + at

20 = 40 +(-4)t

20= 40 -4t

4t = 40-20

4t = 20

t = 20/4

PQ or t = 5s or Answer 

III) Time taken t to cover PR = PQ + QR

But PQ = 5s

And QR =?

Final velocity V= 0m/s

Initial velocity u= 20m/s

Deceleration = Acceleration a = -(-4m/s²)

Using v =u + at

0 = 20 +(4)t

0 = 20 +4t

20-0 = 4t

20= 4t

t = 20/4

t = 5s

Or

s =50m

Let total velocity = initial velocity for the whole journey u = 0m/s

s= ut +½at²

50 = 0 x t + ½ (4)t²

50 = 2t²

50/2 = t²

t = √ 25

t = 5s

Therefore time t taken to cover PR =PQ + PR

t = 5+5 = 10s