Question

A particle is moving with a uniform acceleration its position time t seconds in given in metre by the relation x = 3+4t+5^2. calculate the magnitude of its position at t =5s ​

Answer 1

v = dx/dt

v = d(2+5t+7t²)/dt

v = 5 + 14t

i) initial velocity

= velocity at t = 0

= 5m/s

ii) t = 4

v = 5 + 56

= 61m/s

iii) a = dv/dt

= 14 m/s²

iv) x at t=5

x = 2 + 5×5 + 7×5²

= 2 + 25 + 175

= 202 m

Answer 2

HEYAA!!

Answer:

v = dx/dt

v = d(2+5t+7t²)/dt

v = 5 + 14t

=>initial velocity

= velocity at t = 0

= 5m/s

=> t = 4

t = 4v = 5 + 56

t = 4v = 5 + 56= 61m/s

=> a = dv/dt

a = dv/dt = 14 m/s²

=> x at t=5

at t=5x

= 2 + 5×5 + 7×5²

= 2 + 25 + 175

= 202 m

Hence, it's 202m.

THANKS!! ❤️✌