Question

The motion of a particle is given by y = 40t-st² and x= 40√3 t when x and Y are in metres and 't' in Seconds. Then initial Velocity of the body and angle made by it to the horizontal are

Please answer it​

Answer 1

Answer:

(a) 31²-121-15=0

31²-15t+3t-15=0

31(1-5)+3(1-5)=0

(3t+3)(1-5)=0

Either t = -1 or t=5 Time cannot be negative ..t = 5 seconds.

(b) Position at t=5s x=(5)³-6(5)²-15(5) +40 x=40m

Att = 0 s

x=-60m

Displacement at t = 5s and t = 0s

Xs=-60m

1 =40m

S=-60-40

$=-100m

(c) Acceleration at t = 5s

a = 6(5)-12

a=18m/s²

(a) 31²-121-15=0

31²-15t+3t-15=0

31(1-5)+3(1-5)=0

(3t+3)(1-5)=0

Either t = -1 or t=5 Time cannot be negative ..t = 5 seconds.

(b) Position at t=5s x=(5)³-6(5)²-15(5) +40 x=40m

Att = 0 s

x=-60m

Displacement at t = 5s and t = 0s

Xs=-60m

1 =40m

S=-60-40

$=-100m

(c) Acceleration at t = 5s

a = 6(5)-12

(a) 31²-121-15=0

31²-15t+3t-15=0

31(1-5)+3(1-5)=0

(3t+3)(1-5)=0

Either t = -1 or t=5 Time cannot be negative ..t = 5 seconds.

(b) Position at t=5s x=(5)³-6(5)²-15(5) +40 x=40m

Att = 0 s

x=-60m

Displacement at t = 5s and t = 0s

Xs=-60m

1 =40m

S=-60-40

$=-100m

(c) Acceleration at t = 5s

a = 6(5)-12

(a) 31²-121-15=0

31²-15t+3t-15=0

31(1-5)+3(1-5)=0

(3t+3)(1-5)=0

Either t = -1 or t=5 Time cannot be negative ..t = 5 seconds.

(b) Position at t=5s x=(5)³-6(5)²-15(5) +40 x=40m

Att = 0 s

x=-60m

Displacement at t = 5s and t = 0s

Xs=-60m

1 =40m

S=-60-40

$=-100m

(c) Acceleration at t = 5s

a = 6(5)-12

a = (30-12)

a=18m/s²