Question

The position s of a particle moving along a straight line at time t is given by s = at3+ bt5. Find its
velocity at t=1s.​

Answer 1

Explanation:

$v = \frac{ds}{dt}$

$= \frac{d}{dt}( {at}^{3} + {bt}^{5} )$

$= 3a {t}^{2} + 5b {t}^{4}$

As t is given to be 1s,

v= 3a+5b

Answer 2

Answer :

Given -

• Position of the particle = s = at³ + bt⁵.
• Time (t).

To Find -

• Velocity at t = 1 sec ?

Solution -

Position = s = at³ + bt⁵.

⇒ v = ds/dt

⇒ v = d(at³ + bt⁵)/dt

⇒ v = 3at² + 5bt⁴

Hence, its Instantaneous velocity at t = 1 second :

⇒ 3*a*(1)² + 5*b*(1)⁴

⇒ 3a + 5b

Hence, Instantaneous velocity at t = 1 sec. is : (3a + 5b) m/s.

Additional Information -

• Distance length of the total path travelled by the body.
• Displacement is the shortest length between the initial position and the final position of the body.
• Speed is the distance travelled per unit time.
• Velocity is displacement per unit time.
• Speed and distance is a scalar quantity.
• Velocity and Displacement is a vector quantity.