Question

. A cart is initially moving at 0.5 m/s along a track. The cart comes to rest after traveling 1 m. The experiment is
repeated on the same track, but now the cart is initially moving at 1 m/s. How far does the cart travel before
coming to rest?
(A) 1 m (B) 2 m (C) 3 m (D) 4 m

Answer 1

Answer: Since we know that,

v^2 = u^2 + 2as

Now, the final velocity is zero and deacceleration is constant so

s = (u^2)/2a

Or did directly proportional to u^2

Now u is just doubled so s should become 4 times

s = 4m.

Please mark brainliest if the answer was helpful.

Answer 2

Question

A cart is initially moving at 0.5 m/s along a track. The cart comes to rest after traveling 1 m. The experiment is repeated on the same track, but now the cart is initially moving at 1 m/s. How far does the cart travel before coming to rest?

(A) 1 m

(B) 2 m

(C) 3 m

(D) 4 m

Solution

The cart travels 4m before coming to rest in the second case

Case 1

When,

• Initial Velocity (u) = 0.5 m/s
• Final Velocity (v) = 0 m/s
• Displacement (s) = 1 m
• Acceleration (a) = ?

From the third Kinematic Equation,

$\sf \: {v}^{2} - {u}^{2} = 2as \\ \\ \longrightarrow \: \sf \: 0 {}^{2} - (0.5) {}^{2} = 2(1)a \\ \\ \longrightarrow \: \sf \: - 0.25 = 2a \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \: a = - 0.125 \: {ms}^{ - 2} }}$

Case 2

When,

• Initial Velocity (u) = 1 m/s
• Final Velocity (v) = 0 m/s
• Acceleration (a) = - 0.125 m/s²
• Displacement (s) = ?

From the third Kinematic Equation,

$\longrightarrow \: \sf \: {0}^{2} - {1}^{2} = 2( - 0.125)s \\ \\ \longrightarrow \: \sf \: s = - \dfrac{1}{0.25} \\ \\ \huge{ \longrightarrow \: \boxed{ \boxed{ \sf{s = 4 \: m}}}}$

Option (D) is correct