1. Given situation

A wooden block is on the ground and a pushing force increases linearly with time. So we can write $F(t)=kt$ for some constant $k>0$.

The block is initially at rest.

2. Effect of static friction at the beginning

As long as the applied force is less than or equal to the maximum static friction, $F(t) \le f_{s,\max}=\mu_s N=\mu_s mg,$ the block does not move.

Therefore, for early times, $a=0.$

Let motion start at time $t=t_1$, where $kt_1=\mu_s mg.$ So, $t_1=\frac{\mu_s mg}{k}.$

3. Once the block starts moving

After the block starts sliding, friction becomes kinetic friction: $f_k=\mu_k mg,$ where generally $\mu_k<\mu_s.$

Now net force is $F_{\text{net}}=kt-\mu_k mg.$ Hence acceleration is $a(t)=\frac{kt-\mu_k mg}{m}, \qquad t>t_1.$

This is a straight line increasing with time.

4. Important feature at the instant motion begins

Just before motion starts: $a=0.$

Just after motion starts: $a(t_1^+)=\frac{\mu_s mg-\mu_k mg}{m}=(\mu_s-\mu_k)g>0.$

So acceleration has a sudden jump upward from $0$ to a positive value, because static friction is replaced by smaller kinetic friction.

After that, acceleration increases linearly with time.

5. Shape of the graph

Therefore the acceleration-time graph must be:

• zero initially,
• then a sudden jump at the instant the block starts moving,
• then a straight line with positive slope.

This corresponds to Option D.

6. Comparison with stored answer

Stored correct answer: D

My derived answer: D

So they agree.