Question

1. The x and y coordinates of the particle at any time
are x = 5t - 2t2 and y = 10t respectively, where x
and y are in meters and t in seconds. The
acceleration of the particle at t = 2 s is
[NEET-2017]
(1) O
(2) 5 m/s2
(3) 4 m/s2
(4) 8 m/s2​

Answer 1

Option 3 is right.

• Differentiation of displacement gives velocity.

v = ds/dt

where

ds is the change in displacement.

v is the velocity.

dt is the change in time.

• Differentiation of change in velocity gives the acceleration.

a = dv/dt

where

dv is the change in velocity.

a is the acceleration.

Let's consider the x coordinate:

$x = 5t - 2 {t}^{2} \\ \\ v_x = \frac{dx}{dt} \\ \\ v_x = \frac{d}{dt} (5t - 2 {t}^{2} ) \\ \\ v_x = 5 - (2 \times 2)t \\ \\ v_x = 5 - 4t \\ \\ a_x = \frac{dv}{dt} \\ \\ a_x = \frac{d}{dt} (5 - 4t) \\ \\ a_x = 0 -( 4 \times 1) \\ \\ a_x = - 4 \: \dfrac{m}{ {s}^{2} }$

The acceleration along x coordinate is -4 m/s².

Let's consider y coordinate:

$y = 10t \\ \\ v_y = \frac{dy}{dt} \\ \\ v_y = \frac{d}{dt} (10t) \\ \\ v_y = 10 \times 1 \\ \\ v_y = 10 \: \dfrac{m}{s} \\ \\ a_y = \frac{dv}{dt} \\ \\ a_y = \frac{d}{dt}( 10) \\ \\ a_y = 0$

The acceleration along y coordinate is zero.

Net acceleration :

$a = \sqrt{ {a_x}^{2} + {a_y}^{2} } \\ \\ a = \sqrt{ {( - 4)}^{2} + {(0)}^{2} } \\ \\ a = \sqrt{16} \\ \\ a = 4 \: \frac{m}{ {s}^{2} }$

The net acceleration is 4 m/s².

Answer 2

Option 3 is right.

Differentiation of displacement gives velocity.

v = ds/dt

where

ds is the change in displacement.

v is the velocity.

dt is the change in time.

Differentiation of change in velocity gives the acceleration.

a = dv/dt

where

dv is the change in velocity.

a is the acceleration.

Let's consider the x coordinate:

$x = 5t - 2 {t}^{2} \\ \\ v_x = \frac{dx}{dt} \\ \\ v_x = \frac{d}{dt} (5t - 2 {t}^{2} ) \\ \\ v_x = 5 - (2 \times 2)t \\ \\ v_x = 5 - 4t \\ \\ a_x = \frac{dv}{dt} \\ \\ a_x = \frac{d}{dt} (5 - 4t) \\ \\ a_x = 0 -( 4 \times 1) \\ \\ a_x = - 4 \: \dfrac{m}{ {s}^{2} }$

The acceleration along x coordinate is -4 m/s².

Let's consider y coordinate:

$y = 10t \\ \\ v_y = \frac{dy}{dt} \\ \\ v_y = \frac{d}{dt} (10t) \\ \\ v_y = 10 \times 1 \\ \\ v_y = 10 \: \dfrac{m}{s} \\ \\ a_y = \frac{dv}{dt} \\ \\ a_y = \frac{d}{dt}( 10) \\ \\ a_y = 0$

The acceleration along y coordinate is zero.

Net acceleration :

$a = \sqrt{ {a_x}^{2} + {a_y}^{2} } \\ \\ a = \sqrt{ {( - 4)}^{2} + {(0)}^{2} } \\ \\ a = \sqrt{16} \\ \\ a = 4 \: \frac{m}{ {s}^{2} }$

The net acceleration is 4 m/s².