The distance travelled by a particle in time t is given by s = (4t - 4.5t^2 + 5t

Question

The distance travelled by a particle in time t is given by s = (4t - 4.5t^2 + 5t^3), Find (a) The average speed and acceleration of particle during the time interval 0 to 5s. (b) The instantaneous speed and acceleration of particle at t = 5s. A table clock has its minute hand 5cm long. Find the average velocity of the tip of minute hand (a) Between 6.00am to 6.30am, and (b) Between 6.00am to 6.30pm. A car starts with initial velocity of 30m/s and it accelerates uniformly at the rate 5m/s^2. (a) Find distance travelled in 5s. (b) How much time it will take to reach velocity 45m/s? A ball is kicked at 35m/s at an angle of 45 degree from ground. (a) Find maximum height attained by ball. (b) Velocity at maximum height (c) Time to land (d) How far away does it land

Explanation / Answer

Q1.

s = (4t-4.5t^2+5t^3)

a) v_av = s/t
=>v_av = (4t-4.5t^2+5t^3)/t
=> v_av = (4(0.5)-4.5(0.5)^2+5(0.5)^3)/0.5
= 3 m/s

a = v/t = 3/0.5 = 6 m/s^2

b)
velocity is derivative of position, instantaneous speed is

v = ds/dt
=> v = 15t^2-9t+4
=> v = 15(5)^2-9(5)+4
=> v = 334 m/s

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